CBSE Class 9 Mathematics Polynomials Assignment Set 03

Question. Check, whether 1 is the zero of the polynomial \( 9x^3 - 5x + 20 \).
Answer: Let the given polynomial be \( p(x) = 9x^3 - 5x + 20 \).
To check if 1 is a zero, substitute \( x = 1 \) into the polynomial:
\( p(1) = 9(1)^3 - 5(1) + 20 \)
\( \implies p(1) = 9(1) - 5 + 20 \)
\( \implies p(1) = 9 - 5 + 20 \)
\( \implies p(1) = 24 \)
Since \( p(1) \neq 0 \), 1 is not a zero of the polynomial.
In simple words: We put 1 in place of x in the polynomial. Since the final answer is 24 and not 0, 1 is not a zero of this expression.
Exam Tip: To check if a number is a zero of a polynomial, substitute it for the variable. If the final value is zero, it is a zero; otherwise, it is not.

Question. Factorise \( \frac{2x^2}{8} - \frac{2y^2}{8} \).
Answer: The expression can be simplified as:
\( \frac{2x^2}{8} - \frac{2y^2}{8} = \frac{x^2}{4} - \frac{y^2}{4} \)
\( \implies \frac{x^2}{4} - \frac{y^2}{4} = \left(\frac{x}{2}\right)^2 - \left(\frac{y}{2}\right)^2 \)
Using the identity \( a^2 - b^2 = (a - b)(a + b) \):
\( \implies \left(\frac{x}{2} - \frac{y}{2}\right)\left(\frac{x}{2} + \frac{y}{2}\right) = \frac{1}{4}(x - y)(x + y) \)
In simple words: First simplify the fractions by dividing the numbers. Then, use the identity of difference of squares to write it in factored form.

Exam Tip: Always look for common numerical factors to simplify the algebraic terms before using standard identities.

Question. Show that \( x = 1 \) is a zero of the polynomial \( 3x^3 - 4x^2 + 8x - 7 \).
Answer: Let \( p(x) = 3x^3 - 4x^2 + 8x - 7 \).
Substitute \( x = 1 \) into the polynomial:
\( p(1) = 3(1)^3 - 4(1)^2 + 8(1) - 7 \)
\( \implies p(1) = 3(1) - 4(1) + 8 - 7 \)
\( \implies p(1) = 3 - 4 + 8 - 7 \)
\( \implies p(1) = 0 \)
Since the value of the polynomial is 0 at \( x = 1 \), it is shown that \( x = 1 \) is a zero of the polynomial.
In simple words: Putting 1 instead of x gives a final value of 0. This confirms that 1 is indeed a zero of the polynomial.

Exam Tip: Write down each calculation step clearly to avoid simple arithmetic mistakes, especially when dealing with negative coefficients.

Question. Is \( \left(y^2\right)^{\frac{1}{2}} + 2\sqrt{3} \) a polynomial?
Answer: Let us simplify the given expression:
\( \left(y^2\right)^{\frac{1}{2}} + 2\sqrt{3} = y^{2 \times \frac{1}{2}} + 2\sqrt{3} \)
\( \implies y + 2\sqrt{3} \)
In this simplified expression, the exponent of the variable \( y \) is 1, which is a whole number. Since all exponents of the variables are non-negative integers, the expression is a polynomial.
In simple words: After simplifying the power, we get \( y + 2\sqrt{3} \). Since the variable has a positive whole-number exponent, it is a polynomial.

Exam Tip: Always simplify the powers of variables first. The presence of irrational constants like \( 2\sqrt{3} \) does not affect whether an expression is a polynomial.

Question. Check whether \( \frac{3}{\sqrt{x}} - \sqrt{2}x \) is a polynomial.
Answer: Let us write the given expression in terms of exponents:
\( \frac{3}{\sqrt{x}} - \sqrt{2}x = 3x^{-\frac{1}{2}} - \sqrt{2}x \)
A polynomial cannot have variables with fractional or negative exponents. Since the exponent of \( x \) in the first term is \( -\frac{1}{2} \), which is not a whole number, the expression is not a polynomial.
In simple words: The term \( \frac{3}{\sqrt{x}} \) has a square root of x in the denominator, meaning the exponent is negative and fractional. Therefore, it is not a polynomial.

Exam Tip: A polynomial must only contain variables raised to non-negative integer powers (whole numbers). Watch out for variables in denominators or under radicals.

Question. Find the degree of polynomial \( 30x^5 - 15x^2 + 40 \).
Answer: The degree of a polynomial is the highest power of the variable present in it. In \( 30x^5 - 15x^2 + 40 \), the term with the highest power of \( x \) is \( 30x^5 \), where the exponent is 5. Thus, the degree of the polynomial is 5.
In simple words: The highest power of the variable x in this expression is 5, so the degree is 5.

Exam Tip: The degree is determined solely by the highest exponent of the variable, regardless of the values of the coefficients.

Question. Find the degree of the polynomial \( 4x + 5 \).
Answer: The given polynomial is \( 4x + 5 \), which can be written as \( 4x^1 + 5x^0 \). The highest exponent of the variable \( x \) is 1. Therefore, the degree of the polynomial is 1.
In simple words: Since x has no visible power, its power is 1, which is the highest in the expression. So, the degree is 1.

Exam Tip: Linear polynomials always have a degree of 1, and constant terms have a degree of 0.

Question. If \( a + \frac{1}{a} = b \), then find the value of \( a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \).
Answer: We use the algebraic identity for the cube of a binomial:
\( \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3a\left(\frac{1}{a}\right)\left(a + \frac{1}{a}\right) \)
\( \implies \left(a + \frac{1}{a}\right)^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
We are given that \( a + \frac{1}{a} = b \). Substituting this value into the equation:
\( \implies b^3 = a^3 + \frac{1}{a^3} + 3\left(a + \frac{1}{a}\right) \)
Thus, the value of the given expression is \( b^3 \).
In simple words: The given expression is exactly the expanded form of \( \left(a + \frac{1}{a}\right)^3 \). Since \( a + \frac{1}{a} \) is equal to b, the answer is just \( b^3 \).

Exam Tip: Recognizing standard expansions like \( (x+y)^3 \) directly saves time and prevents algebraic errors during examinations.

Question. Find the remainder when \( 6x^3 - 5x^2 + 2x - 9 \) is divided by \( (x - 1) \).
Answer: Let \( p(x) = 6x^3 - 5x^2 + 2x - 9 \).
According to the Remainder Theorem, when a polynomial \( p(x) \) is divided by \( (x - c) \), the remainder is \( p(c) \). Here, \( c = 1 \).
Substitute \( x = 1 \) into \( p(x) \):
\( p(1) = 6(1)^3 - 5(1)^2 + 2(1) - 9 \)
\( \implies p(1) = 6(1) - 5(1) + 2 - 9 \)
\( \implies p(1) = 6 - 5 + 2 - 9 \)
\( \implies p(1) = -6 \)
Therefore, the remainder is \( -6 \).
In simple words: According to the remainder theorem, we just put 1 in place of x in the expression. Working out the arithmetic gives -6 as the remainder.

Exam Tip: Using the Remainder Theorem is much faster and less error-prone than performing full polynomial long division for linear divisors.

Question. Find the zeroes of the polynomial \( 6x^2 - 3 \).
Answer: To find the zeroes, set the polynomial equal to 0:
\( 6x^2 - 3 = 0 \)
\( \implies 6x^2 = 3 \)
\( \implies x^2 = \frac{3}{6} \)
\( \implies x^2 = \frac{1}{2} \)
Taking the square root on both sides:
\( \implies x = \pm\frac{1}{\sqrt{2}} \)
Hence, the zeroes of the polynomial are \( \frac{1}{\sqrt{2}} \) and \( -\frac{1}{\sqrt{2}} \).
In simple words: We solve \( 6x^2 - 3 = 0 \). Simplifying this gives \( x^2 = \frac{1}{2} \), which means x can be either \( \frac{1}{\sqrt{2}} \) or \( -\frac{1}{\sqrt{2}} \).

Exam Tip: Remember to include both the positive and negative square roots when solving equations of the form \( x^2 = c \).

Question. Find the zeroes of the quadratic polynomial \( x^2 - 4x + 3 \).
Answer: To find the zeroes, we set the quadratic polynomial to 0:
\( x^2 - 4x + 3 = 0 \)
We factorise by splitting the middle term:
\( x^2 - 3x - x + 3 = 0 \)
\( \implies x(x - 3) - 1(x - 3) = 0 \)
\( \implies (x - 3)(x - 1) = 0 \)
Setting each factor to 0:
\( x - 3 = 0 \implies x = 3 \)
\( x - 1 = 0 \implies x = 1 \)
Therefore, the zeroes of the quadratic polynomial are 1 and 3.
In simple words: We split the middle term to factorise the quadratic expression. This gives us factors of \( (x-3) \) and \( (x-1) \), which lead to zeroes at 1 and 3.

Exam Tip: Always verify your zeroes by substituting them back into the original quadratic equation to see if they yield zero.

Question. Find the zeroes of the quadratic polynomial \( x^2 + 7x + 10 \).
Answer: Set the polynomial equal to 0:
\( x^2 + 7x + 10 = 0 \)
Factorise by splitting the middle term \( 7x \) into \( 5x \) and \( 2x \):
\( x^2 + 5x + 2x + 10 = 0 \)
\( \implies x(x + 5) + 2(x + 5) = 0 \)
\( \implies (x + 5)(x + 2) = 0 \)
Setting each factor to 0:
\( x + 5 = 0 \implies x = -5 \)
\( x + 2 = 0 \implies x = -2 \)
The zeroes of the quadratic polynomial are -2 and -5.
In simple words: We factorise the equation into \( (x+5)(x+2) = 0 \). Solving this gives two zeroes: -2 and -5.

Exam Tip: Pay close attention to the signs of the zeroes; if both terms in the factors are positive, the zeroes will be negative.

Question. Given a polynomial \( p(x) \). The graph of \( y = p(x) \) intersects the \( x \)-axis at three points. Find the number of zeroes of \( p(x) \).
Answer: The number of zeroes of a polynomial \( p(x) \) is equal to the number of points where its graph \( y = p(x) \) intersects the \( x \)-axis. Since the graph intersects the \( x \)-axis at three distinct points, the number of zeroes of \( p(x) \) is 3.
In simple words: Every time a graph crosses the x-axis, it represents a zero. Since this graph crosses three times, it has exactly 3 zeroes.

Exam Tip: Remember that only real zeroes correspond to intersection points on the \( x \)-axis of a real coordinate plane.

Question. Find the zeroes of polynomial \( 2x^2 - 8 \).
Answer: To find the zeroes, we solve for \( 2x^2 - 8 = 0 \):
\( 2(x^2 - 4) = 0 \)
\( \implies x^2 - 4 = 0 \)
\( \implies (x - 2)(x + 2) = 0 \)
Setting each factor to 0:
\( x - 2 = 0 \implies x = 2 \)
\( x + 2 = 0 \implies x = -2 \)
Hence, the zeroes of the polynomial are 2 and -2.
In simple words: We factor out the 2, leaving \( x^2 - 4 = 0 \). This simplifies to \( (x-2)(x+2) = 0 \), which gives zeroes of 2 and -2.

Exam Tip: Factoring out the greatest common factor (GCF) first makes working with quadratic terms much simpler.

Question. Using a suitable identity, factorise the following expressions:
(i) \( \frac{x^2}{4} - \frac{y^2}{4} \)
(ii) \( x^2 - \frac{y^2}{100} \)
Answer: We use the difference of squares identity, \( a^2 - b^2 = (a - b)(a + b) \), for both parts:
(i) \( \frac{x^2}{4} - \frac{y^2}{4} = \left(\frac{x}{2}\right)^2 - \left(\frac{y}{2}\right)^2 \)
\( \implies \left(\frac{x}{2} - \frac{y}{2}\right)\left(\frac{x}{2} + \frac{y}{2}\right) = \frac{1}{4}(x - y)(x + y) \)

(ii) \( x^2 - \frac{y^2}{100} = x^2 - \left(\frac{y}{10}\right)^2 \)
\( \implies \left(x - \frac{y}{10}\right)\left(x + \frac{y}{10}\right) \)
In simple words: We write each expression as a difference of two squares and apply the formula \( a^2 - b^2 = (a-b)(a+b) \) to find the factors.

Exam Tip: Always express each term as a complete perfect square, such as writing \( \frac{y^2}{100} \) as \( \left(\frac{y}{10}\right)^2 \), before applying the identity.

Question. Factorise: \( 4y^2 - 4y + 1 \).
Answer: We can rewrite this expression to match the perfect square trinomial identity \( a^2 - 2ab + b^2 = (a - b)^2 \):
\( 4y^2 - 4y + 1 = (2y)^2 - 2(2y)(1) + (1)^2 \)
Here, \( a = 2y \) and \( b = 1 \).
\( \implies (2y - 1)^2 \)
Thus, the factorised form is \( (2y - 1)(2y - 1) \).
In simple words: This expression follows the pattern \( (a-b)^2 \). We can write it as \( (2y-1)^2 \).

Exam Tip: Look closely at the first and last terms to see if they are perfect squares, which often signals a perfect square trinomial.

Question. Show that \( x = 1 \) is a zero of the polynomial \( 2x^3 - 3x^2 + 7x - 6 \).
Answer: Let \( p(x) = 2x^3 - 3x^2 + 7x - 6 \).
Substitute \( x = 1 \) into the polynomial expression:
\( p(1) = 2(1)^3 - 3(1)^2 + 7(1) - 6 \)
\( \implies p(1) = 2(1) - 3(1) + 7 - 6 \)
\( \implies p(1) = 2 - 3 + 7 - 6 \)
\( \implies p(1) = 0 \)
Since evaluating the polynomial at \( x = 1 \) gives 0, \( x = 1 \) is indeed a zero of \( p(x) \).
In simple words: Substituting 1 for x results in a total value of 0, which proves that 1 is a zero of this expression.

Exam Tip: Be careful with the arithmetic when adding and subtracting positive and negative terms together.

Question. Give one example each of a binomial of degree 35 and a monomial of degree 100.
Answer:
1. A binomial has exactly two terms, and its degree is the highest exponent of the variable. An example of a binomial of degree 35 is:
\( x^{35} - 9 \)
2. A monomial has exactly one term, and its degree is the exponent of the variable. An example of a monomial of degree 100 is:
\( 5y^{100} \)
In simple words: A binomial has two terms, so \( x^{35} - 9 \) works. A monomial has only one term, so \( 5y^{100} \) works.

Exam Tip: Ensure that your binomial has exactly two terms with the highest exponent being 35, and your monomial has exactly one term with an exponent of 100.

Question. Classify the following polynomials as linear, quadratic, cubic & bi-quadratic polynomials:
(i) \( x + x^2 + 4 \)
(ii) \( 3x - 2 \)
(iii) \( 3y \)
(iv) \( 7t^4 + 4t^3 + 3t - 2 \)
Answer: We classify each polynomial based on its degree (the highest power of its variable):
(i) \( x + x^2 + 4 \): The highest power is 2, so the degree is 2. This is a quadratic polynomial.
(ii) \( 3x - 2 \): The highest power is 1, so the degree is 1. This is a linear polynomial.
(iii) \( 3y \): The highest power is 1, so the degree is 1. This is a linear polynomial.
(iv) \( 7t^4 + 4t^3 + 3t - 2 \): The highest power is 4, so the degree is 4. This is a bi-quadratic polynomial.
In simple words: Polynomials are named by their highest power. Power 1 is linear, power 2 is quadratic, and power 4 is bi-quadratic.

Exam Tip: Remember that a cubic polynomial has a degree of 3, while a bi-quadratic polynomial has a degree of 4.

Question. Write the degrees of each of the following polynomials:
(i) \( 7x^3 + 4x^2 - 3x + 12 \)
(ii) \( 12 - x + 2x^3 \)
(iii) \( 5x - \sqrt{2} \)
(iv) \( 7 \)
Answer: The degree is the highest exponent of the variable in each polynomial:
(i) For \( 7x^3 + 4x^2 - 3x + 12 \), the term with the highest power of \( x \) is \( 7x^3 \). Thus, the degree is 3.
(ii) For \( 12 - x + 2x^3 \), the term with the highest power of \( x \) is \( 2x^3 \). Thus, the degree is 3.
(iii) For \( 5x - \sqrt{2} \), the term with the highest power of \( x \) is \( 5x^1 \). Thus, the degree is 1.
(iv) For \( 7 \), this is a non-zero constant polynomial which can be written as \( 7x^0 \). Thus, the degree is 0.
In simple words: The degree is simply the biggest power of the variable in the expression. Constants have a degree of 0.

Exam Tip: Do not confuse constant polynomials with the zero polynomial; a non-zero constant has degree 0, while the degree of the zero polynomial is undefined.

Question. Write the coefficient of \( x^2 \) in each of the following:
(1) \( 17 - 2x + 7x^2 \)
(2) \( 9 - 12x + x^3 \)
(3) \( \frac{\pi}{6}x^2 - 3x + 4 \)
(4) \( \sqrt{3}x - 7 \)
Answer: The coefficient is the numerical factor multiplying the term \( x^2 \):
(1) In the expression \( 17 - 2x + 7x^2 \), the term containing \( x^2 \) is \( 7x^2 \). Its coefficient is 7.
(2) In the expression \( 9 - 12x + x^3 \), there is no term with \( x^2 \). This can be written with a term \( 0x^2 \). Thus, the coefficient of \( x^2 \) is 0.
(3) In the expression \( \frac{\pi}{6}x^2 - 3x + 4 \), the term containing \( x^2 \) is \( \frac{\pi}{6}x^2 \). Its coefficient is \( \frac{\pi}{6} \).
(4) In the expression \( \sqrt{3}x - 7 \), there is no term with \( x^2 \). Thus, the coefficient of \( x^2 \) is 0.
In simple words: We look for the number multiplied by \( x^2 \). If there is no \( x^2 \) term at all, then the coefficient is 0.

Exam Tip: If a variable term is absent, its coefficient in that polynomial is always considered to be 0.

Question. Find the zeroes of the quadratic polynomial \( x^2 - 4x + 3 \).
Answer: Set the polynomial equal to 0 to solve for its roots:
\( x^2 - 4x + 3 = 0 \)
We split the middle term \( -4x \) as \( -3x - x \):
\( x^2 - 3x - x + 3 = 0 \)
\( \implies x(x - 3) - 1(x - 3) = 0 \)
\( \implies (x - 3)(x - 1) = 0 \)
This gives us:
\( x - 3 = 0 \implies x = 3 \)
\( x - 1 = 0 \implies x = 1 \)
Thus, the zeroes of the quadratic polynomial are 1 and 3.
In simple words: By factorising the expression, we get the two factors \( (x-3) \) and \( (x-1) \). Setting these to zero gives our solutions, which are 1 and 3.

Exam Tip: Ensure that the product of the two split constants equals the constant term and their sum equals the coefficient of the middle term.

Question. Resolve into factors: \( 27x^3 + y^3 + z^3 - 9xyz \).
Answer: We use the algebraic identity:
\( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) \)
The given expression is \( 27x^3 + y^3 + z^3 - 9xyz \).
We can rewrite this expression as:
\( (3x)^3 + y^3 + z^3 - 3(3x)(y)(z) \)
Here, \( a = 3x \), \( b = y \), and \( c = z \).
Applying the identity:
\( \implies (3x + y + z)((3x)^2 + y^2 + z^2 - (3x)(y) - y(z) - z(3x)) \)
\( \implies (3x + y + z)(9x^2 + y^2 + z^2 - 3xy - yz - 3zx) \)
In simple words: We rewrite the term \( 27x^3 \) as \( (3x)^3 \) and then use a standard algebraic identity for three cubed variables to factorise it.

Exam Tip: Be careful when squaring \( 3x \); the term must become \( 9x^2 \), not \( 3x^2 \).

Question. Factorise: \( 64a^3 - 27b^3 - 144a^2b + 108ab^2 \).
Answer: We can rewrite the expression to see if it matches the cubic expansion identity \( (x - y)^3 = x^3 - y^3 - 3x^2y + 3xy^2 \):
\( 64a^3 = (4a)^3 \)
\( 27b^3 = (3b)^3 \)
Let us test the remaining terms with \( x = 4a \) and \( y = 3b \):
\( -3x^2y = -3(4a)^2(3b) = -3(16a^2)(3b) = -144a^2b \)
\( 3xy^2 = 3(4a)(3b)^2 = 3(4a)(9b^2) = 108ab^2 \)
Since both match our given expression, we can write:
\( 64a^3 - 27b^3 - 144a^2b + 108ab^2 = (4a - 3b)^3 \)
This can be written as \( (4a - 3b)(4a - 3b)(4a - 3b) \).
In simple words: The expression is a perfect cube of \( 4a - 3b \) because it fits the formula for \( (x-y)^3 \).

Exam Tip: Identifying perfect cubes in the leading and trailing terms is a strong hint that the expression might simplify to \( (x \pm y)^3 \).

Question. If \( p = 2 - a \), then show that \( a^3 + 6ap + p^3 - 8 = 0 \).
Answer: We are given the relation:
\( p = 2 - a \)
Rearranging the terms, we get:
\( a + p = 2 \)
Now, we cube both sides of this equation:
\( (a + p)^3 = 2^3 \)
Using the identity \( (x + y)^3 = x^3 + y^3 + 3xy(x + y) \):
\( \implies a^3 + p^3 + 3ap(a + p) = 8 \)
Substitute \( a + p = 2 \) back into the equation:
\( \implies a^3 + p^3 + 3ap(2) = 8 \)
\( \implies a^3 + p^3 + 6ap = 8 \)
Subtracting 8 from both sides:
\( \implies a^3 + 6ap + p^3 - 8 = 0 \)
Hence, the given relation is shown.
In simple words: We rewrite the equation as \( a+p=2 \) and cube both sides. Substituting 2 back in for \( a+p \) gives the exact expression we need to prove.

Exam Tip: Substituting the original sum \( a + p \) back into the expanded identity is the key step to solving this type of proof question.

Question. If \( x + y + z = 6 \), \( xy + yz + zx = 11 \). Find the value of \( x^2 + y^2 + z^2 \).
Answer: We use the algebraic identity for the square of a trinomial:
\( (x + y + z)^2 = x^2 + y^2 + z^2 + 2(xy + yz + zx) \)
We substitute the given values into this identity:
\( (6)^2 = x^2 + y^2 + z^2 + 2(11) \)
\( \implies 36 = x^2 + y^2 + z^2 + 22 \)
Isolating the term \( x^2 + y^2 + z^2 \):
\( \implies x^2 + y^2 + z^2 = 36 - 22 \)
\( \implies x^2 + y^2 + z^2 = 14 \)
Thus, the value of the expression is 14.
In simple words: We plug the given numbers into the formula for \( (x+y+z)^2 \). We square 6 to get 36, then subtract 22 to get 14.

Exam Tip: Memorize the trinomial square expansion as it is frequently tested in algebraic evaluation problems.

Question. Factorise:
(i) \( 27y^3 + 125z^3 \)
(ii) \( 64m^3 - 343n^3 \)
Answer: We apply the sum and difference of cubes identities:
(i) Using \( a^3 + b^3 = (a + b)(a^2 - ab + b^2) \):
\( 27y^3 + 125z^3 = (3y)^3 + (5z)^3 \)
\( \implies (3y + 5z)((3y)^2 - (3y)(5z) + (5z)^2) \)
\( \implies (3y + 5z)(9y^2 - 15yz + 25z^2) \)

(ii) Using \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \):
\( 64m^3 - 343n^3 = (4m)^3 - (7n)^3 \)
\( \implies (4m - 7n)((4m)^2 + (4m)(7n) + (7n)^2) \)
\( \implies (4m - 7n)(16m^2 + 28mn + 49n^2) \)
In simple words: We rewrite both parts using their cube roots and then use the standard formulas for the sum and difference of cubes to split them into factors.

Exam Tip: Be extremely careful with signs in these formulas; the sum of cubes has a minus in the trinomial, while the difference of cubes has a plus.

Question. Factorise: \( 27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4}p \).
Answer: We can rewrite this expression to see if it fits the cubic identity \( (a - b)^3 = a^3 - b^3 - 3a^2b + 3ab^2 \):
\( 27p^3 = (3p)^3 \)
\( \frac{1}{216} = \left(\frac{1}{6}\right)^3 \)
Let \( a = 3p \) and \( b = \frac{1}{6} \). Checking the middle terms:
\( -3a^2b = -3(3p)^2\left(\frac{1}{6}\right) = -3(9p^2)\left(\frac{1}{6}\right) = -\frac{9}{2}p^2 \)
\( 3ab^2 = 3(3p)\left(\frac{1}{6}\right)^2 = 9p\left(\frac{1}{36}\right) = \frac{1}{4}p \)
Since both terms match our expression, we can write:
\( 27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4}p = \left(3p - \frac{1}{6}\right)^3 \)
This can be written in factored form as \( \left(3p - \frac{1}{6}\right)\left(3p - \frac{1}{6}\right)\left(3p - \frac{1}{6}\right) \).
In simple words: The expression matches the expansion of a perfect cube, which is \( \left(3p - \frac{1}{6}\right)^3 \).

Exam Tip: Whenever fractional terms are present in cubic polynomials, try to express them as perfect cubes of simple unit fractions to find the value of b.

Question. Find the value of \( k \), if \( (x - 1) \) is a factor of the following expression: \( p(x) = kx^2 - \sqrt{2}x + 1 \).
Answer: According to the Factor Theorem, if \( (x - 1) \) is a factor of \( p(x) \), then \( p(1) = 0 \).
Substitute \( x = 1 \) into the polynomial:
\( p(1) = k(1)^2 - \sqrt{2}(1) + 1 \)
\( \implies 0 = k(1) - \sqrt{2} + 1 \)
\( \implies 0 = k - \sqrt{2} + 1 \)
Isolating \( k \):
\( \implies k = \sqrt{2} - 1 \)
Thus, the value of \( k \) is \( \sqrt{2} - 1 \).
In simple words: Since \( x-1 \) is a factor, putting 1 in place of x must give 0. Solving this gives us \( k = \sqrt{2} - 1 \).

Exam Tip: Always make sure to change the sign of the constant term in the divisor \( (x - c) \) when applying the Factor Theorem; here, \( c = 1 \).

Question. Divide the polynomial \( 3x^4 - 4x^3 - 3x - 1 \) by \( x - 1 \).
Answer: We can perform polynomial long division to divide \( 3x^4 - 4x^3 - 3x - 1 \) by \( x - 1 \):
First, write the polynomial including any missing powers with 0 coefficients:
\( 3x^4 - 4x^3 + 0x^2 - 3x - 1 \)
Now, divide step-by-step:
1. Divide the first term: \( \frac{3x^4}{x} = 3x^3 \).
Multiply \( 3x^3(x - 1) = 3x^4 - 3x^3 \).
Subtract to get the remainder: \( -x^3 + 0x^2 - 3x - 1 \).

2. Divide the new first term: \( \frac{-x^3}{x} = -x^2 \).
Multiply \( -x^2(x - 1) = -x^3 + x^2 \).
Subtract to get the remainder: \( -x^2 - 3x - 1 \).

3. Divide the new first term: \( \frac{-x^2}{x} = -x \).
Multiply \( -x(x - 1) = -x^2 + x \).
Subtract to get the remainder: \( -4x - 1 \).

4. Divide the new first term: \( \frac{-4x}{x} = -4 \).
Multiply \( -4(x - 1) = -4x + 4 \).
Subtract to get the final remainder: \( -5 \).

Thus, the quotient is \( 3x^3 - x^2 - x - 4 \) and the remainder is \( -5 \).
In simple words: We divide step-by-step using polynomial long division. This gives us a quotient of \( 3x^3 - x^2 - x - 4 \) with a leftover remainder of -5.

Exam Tip: Be very careful when subtracting negative terms during polynomial division; a common mistake is getting the wrong sign for the remainder.

Question. Check whether \( 7 + 3x \) is a factor of \( 3x^3 + 7x \).
Answer: Let \( p(x) = 3x^3 + 7x \).
To find if \( 7 + 3x \) is a factor, we set \( 7 + 3x = 0 \), which gives the zero \( x = -\frac{7}{3} \).
According to the Factor Theorem, if \( 7 + 3x \) is a factor, then \( p\left(-\frac{7}{3}\right) \) must be 0.
Substitute \( x = -\frac{7}{3} \) into the polynomial:
\( p\left(-\frac{7}{3}\right) = 3\left(-\frac{7}{3}\right)^3 + 7\left(-\frac{7}{3}\right) \)
\( \implies p\left(-\frac{7}{3}\right) = 3\left(-\frac{343}{27}\right) - \frac{49}{3} \)
\( \implies p\left(-\frac{7}{3}\right) = -\frac{343}{9} - \frac{49}{3} \)
Making a common denominator of 9:
\( \implies p\left(-\frac{7}{3}\right) = -\frac{343}{9} - \frac{147}{9} = -\frac{490}{9} \)
Since \( p\left(-\frac{7}{3}\right) \neq 0 \), we conclude that \( 7 + 3x \) is not a factor of \( 3x^3 + 7x \).
In simple words: We find the root of \( 7+3x \), which is \( -\frac{7}{3} \), and plug it into our expression. Since the result is not zero, \( 7+3x \) is not a factor.

Exam Tip: Always fully simplify the fraction when checking factor values to be absolutely certain of the non-zero result.

Question. Find the zero of the polynomial in each of the following cases:
(i) \( h(x) = 2x \)
(ii) \( p(x) = cx + d \), \( c \neq 0 \)
(iii) \( p(x) = ax \), \( a \neq 0 \)
Answer: To find the zero of each polynomial, we set the polynomial to 0 and solve for \( x \):
(i) \( h(x) = 0 \)
\( \implies 2x = 0 \)
\( \implies x = 0 \)
Thus, the zero is 0.

(ii) \( p(x) = 0 \)
\( \implies cx + d = 0 \)
\( \implies cx = -d \)
\( \implies x = -\frac{d}{c} \)
Thus, the zero is \( -\frac{d}{c} \).

(iii) \( p(x) = 0 \)
\( \implies ax = 0 \)
Since \( a \neq 0 \), we divide both sides by \( a \):
\( \implies x = 0 \)
Thus, the zero is 0.
In simple words: To find the zero of a function, we just set the entire expression to 0 and solve for x.

Exam Tip: For any linear polynomial of the form \( px + q \), the zero is always given by the formula \( x = -\frac{q}{p} \).

Question. If \( x = \frac{4}{3} \) is a zero of the polynomial \( f(x) = 2x^3 - 11x^2 + kx - 20 \), find the value of \( k \).
Answer: Since \( x = \frac{4}{3} \) is a zero of \( f(x) \), we have \( f\left(\frac{4}{3}\right) = 0 \).
Substitute \( x = \frac{4}{3} \) into the polynomial equation:
\( 2\left(\frac{4}{3}\right)^3 - 11\left(\frac{4}{3}\right)^2 + k\left(\frac{4}{3}\right) - 20 = 0 \)
\( \implies 2\left(\frac{64}{27}\right) - 11\left(\frac{16}{9}\right) + \frac{4k}{3} - 20 = 0 \)
\( \implies \frac{128}{27} - \frac{176}{9} + \frac{4k}{3} - 20 = 0 \)
Multiply the entire equation by 27 to eliminate the denominators:
\( \implies 128 - 176(3) + 4k(9) - 20(27) = 0 \)
\( \implies 128 - 528 + 36k - 540 = 0 \)
Combine the constant terms:
\( \implies 36k - 940 = 0 \)
\( \implies 36k = 940 \)
\( \implies k = \frac{940}{36} = \frac{235}{9} \)
Thus, the value of \( k \) is \( \frac{235}{9} \).
In simple words: Since \( \frac{4}{3} \) is a zero, putting it in place of x must make the expression equal 0. Working out the fractions tells us that k is \( \frac{235}{9} \).

Exam Tip: Multiplying the entire equation by the lowest common multiple (LCM) of the denominators is a great way to simplify fraction-heavy linear equations.

Question. Identify constant, linear, quadratic & cubic polynomials from the following polynomials:
(i) \( f(x) = 0 \)
(ii) \( g(x) = 2x^3 - 7x + 4 \)
(iii) \( h(x) = -3x + \frac{1}{2} \)
(iv) \( p(x) = 2x^2 - x + 4 \)
(v) \( q(x) = 4x + 3 \)
(vi) \( r(x) = 3x^3 + 4x^2 + 5x - 7 \)
Answer: We classify each polynomial based on its degree:
(i) \( f(x) = 0 \): This is the zero polynomial, which is classified as a constant polynomial.
(ii) \( g(x) = 2x^3 - 7x + 4 \): The highest power of the variable is 3. This is a cubic polynomial.
(iii) \( h(x) = -3x + \frac{1}{2} \): The highest power of the variable is 1. This is a linear polynomial.
(iv) \( p(x) = 2x^2 - x + 4 \): The highest power of the variable is 2. This is a quadratic polynomial.
(v) \( q(x) = 4x + 3 \): The highest power of the variable is 1. This is a linear polynomial.
(vi) \( r(x) = 3x^3 + 4x^2 + 5x - 7 \): The highest power of the variable is 3. This is a cubic polynomial.
In simple words: We look at the highest exponent in each equation. Power 1 is linear, power 2 is quadratic, power 3 is cubic, and 0 is constant.

Exam Tip: Classify polynomials carefully by finding the single highest power of the variable, ignoring other lower-degree terms.

Question. Give possible expressions for the length & breadth of the rectangle whose area is given by \( A = 25a^2 - 35a + 12 \).
Answer: Since Area = Length \( \times \) Breadth, we can find possible expressions for length and breadth by factorising the quadratic expression:
\( A = 25a^2 - 35a + 12 \)
We split the middle term \( -35a \) into two terms whose product is \( 25 \times 12 = 300 \) and whose sum is \( -35 \). These numbers are \( -20 \) and \( -15 \):
\( A = 25a^2 - 20a - 15a + 12 \)
Group the terms to find common factors:
\( \implies A = 5a(5a - 4) - 3(5a - 4) \)
\( \implies A = (5a - 3)(5a - 4) \)
Therefore, possible expressions for the length and breadth of the rectangle are \( (5a - 3) \) and \( (5a - 4) \).
In simple words: We factorise the area expression into two parts. These two factors, \( 5a-3 \) and \( 5a-4 \), represent the possible length and breadth of the rectangle.

Exam Tip: For area factorisation problems, remember that either factor can represent the length or breadth, so both solutions are equally valid.

Question. Factorise the polynomial x3 - 23x2 + 142x - 120.
Answer: Let \( p(x) = x^3 - 23x^2 + 142x - 120 \). By testing divisors of the constant term \( 120 \), we find that \( p(1) = 1^3 - 23(1)^2 + 142(1) - 120 = 1 - 23 + 142 - 120 = 0 \). Hence, by the factor theorem, \( (x - 1) \) is a factor of the polynomial. Dividing \( p(x) \) by \( (x - 1) \), we get the quadratic quotient \( x^2 - 22x + 120 \). Splitting the middle term of this quadratic expression gives \( x^2 - 12x - 10x + 120 = x(x - 12) - 10(x - 12) = (x - 12)(x - 10) \). Therefore, the fully factorised polynomial is \( (x - 1)(x - 10)(x - 12) \).
In simple words: First, find a number that makes the equation equal zero, which is 1. Since 1 works, \( (x-1) \) is a factor, and we can divide to break down the rest into \( (x-10)(x-12) \).
Exam Tip: Always look at the constant term first to test easy values like 1, -1, 2, or -2 for your starting factor.

Question. Factorise:
(i) 12x2 - 7x + 1
(ii) f(x) = 2x2 + 7x + 3
Answer:
(i) For \( 12x^2 - 7x + 1 \), we split the middle term into two numbers whose product is \( 12 \times 1 = 12 \) and whose sum is \( -7 \). These numbers are \( -4 \) and \( -3 \).
\( 12x^2 - 4x - 3x + 1 = 4x(3x - 1) - 1(3x - 1) = (3x - 1)(4x - 1) \).
(ii) For \( 2x^2 + 7x + 3 \), we split the middle term into two parts whose product is \( 2 \times 3 = 6 \) and whose sum is \( 7 \). These numbers are \( 6 \) and \( 1 \).
\( 2x^2 + 6x + x + 3 = 2x(x + 3) + 1(x + 3) = (x + 3)(2x + 1) \).
In simple words: To factorise these quadratic equations, find two numbers that multiply to give the product of the first and last numbers, and add up to the middle number.
Exam Tip: Be careful with signs when grouping terms. For example, in (i), pulling out a negative 1 changes the signs inside the bracket correctly.

Question. Write each of the following expressions, as a product of linear factors, with integer coefficient:
(i) 5x2 + 16x + 3
(ii) 24p2 - 41p + 12
Answer:
(i) Splitting the middle term of \( 5x^2 + 16x + 3 \) using the numbers \( 15 \) and \( 1 \):
\( 5x^2 + 15x + x + 3 = 5x(x + 3) + 1(x + 3) = (x + 3)(5x + 1) \).
(ii) Splitting the middle term of \( 24p^2 - 41p + 12 \) using the numbers \( -32 \) and \( -9 \) (since \( -32 \times -9 = 288 \) and \( -32 - 9 = -41 \)):
\( 24p^2 - 32p - 9p + 12 = 8p(3p - 4) - 3(3p - 4) = (3p - 4)(8p - 3) \).
In simple words: Break down the middle number into two parts so you can group the terms and write the expression as two simple linear factors.
Exam Tip: Check your final linear factors by multiplying them back together to ensure they return the original quadratic expression.

Question. Use the factor theorem to determine whether g(x) is a factor of f(x) in each of the following cases:
(i) f(x) = x3 - 3x2 + 4x - 4, g(x) = x - 2
(ii) f(x) = 2\sqrt{2}x2 + 5x + \sqrt{2}, g(x) = x + \sqrt{2}
Answer:
(i) By the factor theorem, \( g(x) = x - 2 \) is a factor if \( f(2) = 0 \).
Evaluating \( f(2) \):
\( f(2) = 2^3 - 3(2)^2 + 4(2) - 4 = 8 - 12 + 8 - 4 = 0 \).
Since \( f(2) = 0 \), \( g(x) \) is indeed a factor of \( f(x) \).
(ii) For \( g(x) = x + \sqrt{2} \), we check if \( f(-\sqrt{2}) = 0 \).
Evaluating \( f(-\sqrt{2}) \):
\( f(-\sqrt{2}) = 2\sqrt{2}(-\sqrt{2})^2 + 5(-\sqrt{2}) + \sqrt{2} = 2\sqrt{2}(2) - 5\sqrt{2} + \sqrt{2} = 4\sqrt{2} - 5\sqrt{2} + \sqrt{2} = 0 \).
Since \( f(-\sqrt{2}) = 0 \), \( g(x) \) is a factor of \( f(x) \).
In simple words: If you plug the root of g(x) into f(x) and the final answer is zero, then g(x) divides f(x) perfectly with no remainder.
Exam Tip: Remember to switch the sign of the constant in g(x) when finding the value to substitute. For \( x + \sqrt{2} \), you must substitute \( -\sqrt{2} \).

Question. Check whether the following polynomials have (x + 1) as a factor.
(i) x3 + x2 + x + 1
(ii) x3 - x2 - (2 + \sqrt{2})x + \sqrt{2}
Answer:
According to the factor theorem, \( (x + 1) \) is a factor of a polynomial if the value of the polynomial is \( 0 \) at \( x = -1 \).
(i) Substituting \( x = -1 \):
\( (-1)^3 + (-1)^2 + (-1) + 1 = -1 + 1 - 1 + 1 = 0 \).
Since the value is \( 0 \), \( (x + 1) \) is a factor.
(ii) Substituting \( x = -1 \):
\( (-1)^3 - (-1)^2 - (2 + \sqrt{2})(-1) + \sqrt{2} = -1 - 1 + (2 + \sqrt{2}) + \sqrt{2} = -2 + 2 + 2\sqrt{2} = 2\sqrt{2} \).
Since \( 2\sqrt{2} \neq 0 \), \( (x + 1) \) is not a factor.
In simple words: Replace x with -1 in both equations. The first one equals zero, so it has \( (x+1) \) as a factor, but the second one does not.
Exam Tip: Double-check sign distributions carefully, especially when multiplying a negative number like -1 into parentheses containing a negative sign.

Question. Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1
(ii) x - 1/2
Answer:
Let \( p(x) = x^3 + 3x^2 + 3x + 1 \). According to the remainder theorem, the remainder when \( p(x) \) is divided by \( x - a \) is \( p(a) \).
(i) For divisor \( x + 1 \), the remainder is \( p(-1) \):
\( p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1 = -1 + 3 - 3 + 1 = 0 \).
(ii) For divisor \( x - \frac{1}{2} \), the remainder is \( p\left(\frac{1}{2}\right) \):
\( p\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) + 1 = \frac{1}{8} + \frac{3}{4} + \frac{3}{2} + 1 = \frac{1 + 6 + 12 + 8}{8} = \frac{27}{8} \).
In simple words: The remainder theorem lets us find the leftover remainder without doing long division. Just substitute the root of the divisor into the main equation.
Exam Tip: When working with fractions in remainder questions, always find a common denominator carefully before adding the terms.

Question. Which of the following expressions are polynomial and which are not? State reasons for your answer.
(1) 3x2 - 4x + 15
(2) y2 + 2\sqrt{3}
(3) 3\sqrt{x} - 2\sqrt{x}
Answer:
An expression is a polynomial if all exponents of its variable(s) are non-negative integers.
(1) \( 3x^2 - 4x + 15 \) is a polynomial because the powers of \( x \) are \( 2 \) and \( 1 \), which are both non-negative integers.
(2) \( y^2 + 2\sqrt{3} \) is a polynomial since the variable \( y \) has an exponent of \( 2 \) (a non-negative integer). The constant term containing a square root does not affect polynomial status.
(3) \( 3\sqrt{x} - 2\sqrt{x} = \sqrt{x} = x^{1/2} \) is not a polynomial because the exponent of \( x \) is \( \frac{1}{2} \), which is not an integer.
In simple words: A polynomial cannot have a variable under a square root or as a fraction in its final simplified form. Only the constant numbers can have roots.
Exam Tip: Always simplify the expression first before deciding if it is a polynomial or not.

Question. If f(x) = 2x3 - 13x2 + 17x + 12 , find
(i) f(2)
(ii) f(-3)
Answer:
(i) To find \( f(2) \), substitute \( x = 2 \) into the polynomial:
\( f(2) = 2(2)^3 - 13(2)^2 + 17(2) + 12 = 2(8) - 13(4) + 34 + 12 = 16 - 52 + 34 + 12 = 10 \).
(ii) To find \( f(-3) \), substitute \( x = -3 \) into the polynomial:
\( f(-3) = 2(-3)^3 - 13(-3)^2 + 17(-3) + 12 = 2(-27) - 13(9) - 51 + 12 = -54 - 117 - 51 + 12 = -210 \).
In simple words: Substitute the given numbers in place of x and calculate the final arithmetic value step-by-step.
Exam Tip: Be very careful when raising a negative number to an odd power, as it keeps its negative sign (e.g., \( (-3)^3 = -27 \)).