CBSE Class 9 Mathematics Polynomials Assignment Set 11

CLASS IX - ASSIGNMENT NO. 1
POLYNOMIALS

Q1. Which of the following expressions are Polynomials in One Variable
(a) \(ax^2 + bx + c\)
(b) \(2x^3 - 5x + 6\)
(c) \(x - \frac{1}{x}\)
(d) \(\sqrt{3}x^2 + 5\sqrt{3}x + \sqrt{6}\)
(e) \(3\sqrt{x} + x\sqrt{3}\)
Answer: (a), (b), (d)
A polynomial in one variable is an expression containing only one variable with non-negative integer exponents, and where the coefficients are constants.
(a) \(ax^2 + bx + c\) is a polynomial in one variable \(x\) (assuming \(a, b, c\) are constants).
(b) \(2x^3 - 5x + 6\) is a polynomial in one variable \(x\) with whole number exponents \(3\) and \(1\).
(c) \(x - \frac{1}{x} = x - x^{-1}\) is not a polynomial because the exponent of \(x\) in the second term is \(-1\), which is a negative integer.
(d) \(\sqrt{3}x^2 + 5\sqrt{3}x + \sqrt{6}\) is a polynomial in one variable \(x\) because the variable has whole number exponents (\(2\) and \(1\)), and the coefficients are real numbers.
(e) \(3\sqrt{x} + x\sqrt{3} = 3x^{\frac{1}{2}} + \sqrt{3}x\) is not a polynomial because the exponent of \(x\) in the first term is \(\frac{1}{2}\), which is not a whole number.
In simple words: Expressions are polynomials if the powers of the variable are whole numbers and there is only one letter variable.

Exam Tip: Always check that the variable is not inside a square root and does not appear in any denominator.

Q2. Find the Coefficient of \(x^3\) in \(4x^3 - 3x^2 + 5x + 6\) and the Coefficient of \(x\) in \(4x^2 + 6\)
Answer: 4, 0
For the polynomial \(4x^3 - 3x^2 + 5x + 6\), the term containing \(x^3\) is \(4x^3\). Its coefficient is \(4\).
For the polynomial \(4x^2 + 6\), which can be written as \(4x^2 + 0x + 6\), the term containing \(x\) is absent, meaning its coefficient is \(0\).
In simple words: The coefficient is simply the number multiplied by the specific power of the variable.

Exam Tip: If a term is missing from an algebraic expression, its coefficient is always taken as zero.

Q3. Rewrite the following Polynomial in standard form and write the degree
(a) \(-2x + 5 + 6x^2 - 7x^3\)
(b) \(x^4 - 5 + 6x + 7x^3\)
Answer:
(a) Standard form: \(-7x^3 + 6x^2 - 2x + 5\), Degree = \(3\)
(b) Standard form: \(x^4 + 7x^3 + 6x - 5\), Degree = \(4\)
In simple words: Standard form means arranging the terms from the highest power of the variable to the lowest power. The degree is the highest power.

Exam Tip: Always arrange the powers in descending order to easily identify the degree of the polynomial.

Q4. Which of the following Polynomials are Monomials, binomials, trinomials
(a) \(x^3 + 5x^2 + 5x + 3\)
(b) \(x^4\)
(c) \(x^5 + 5x^4 + 3\)
(d) \(x^2 + x - x + 6\)
Answer:
(a) \(x^3 + 5x^2 + 5x + 3\) has 4 terms, so it is a polynomial with 4 terms (Quadrinomial / general polynomial).
(b) \(x^4\) has 1 term, so it is a Monomial.
(c) \(x^5 + 5x^4 + 3\) has 3 terms, so it is a Trinomial.
(d) \(x^2 + x - x + 6 = x^2 + 6\) has 2 terms, so it is a Binomial.
In simple words: Count the number of terms separated by plus or minus signs after simplifying: 1 term = monomial, 2 terms = binomial, 3 terms = trinomial.

Exam Tip: Always simplify the expression completely by combining like terms before naming its type.

Q5. Find the Value of the Polynomial \(p(x)\) at \(x = -1\)
\(p(x) = 2x^3 + 6x^2 + 5x - 3\)
Answer: -4
Substitute \(x = -1\) into the given polynomial:
\(p(-1) = 2(-1)^3 + 6(-1)^2 + 5(-1) - 3\)
\(p(-1) = 2(-1) + 6(1) - 5 - 3\)
\(p(-1) = -2 + 6 - 5 - 3\)
\(p(-1) = 4 - 5 - 3 = -1 - 3 = -4\)
In simple words: Replace every \(x\) with \(-1\) and evaluate the arithmetic expression following BODMAS.

Exam Tip: Be very careful with negative signs when raising numbers to odd and even powers.

Q6. Check whether \(-3\) and \(3\) are the zeroes of the polynomial \(p(x) = x + 3\)
Answer: \(-3\) is a zero, \(3\) is not a zero
For \(x = -3\):
\(p(-3) = -3 + 3 = 0\)
Since \(p(-3) = 0\), \(-3\) is a zero of the polynomial.
For \(x = 3\):
\(p(3) = 3 + 3 = 6\)
Since \(p(3) \neq 0\), \(3\) is not a zero of the polynomial.
In simple words: A number is a zero of a polynomial if substituting it makes the polynomial equal to zero.

Exam Tip: Always substitute the given value and write the final conclusion explicitly.

Q7. Verify whether the indicated numbers are zeroes of the polynomial or not
(a) \(p(x) = 3x - 2\), \(x = \frac{2}{3}\)
(b) \(p(x) = x^2 - 9\), \(x = 3, -3\)
(c) \(p(x) = 5x - \pi\), \(x = \pi\)
Answer: (a) Yes, (b) Yes, (c) No
(a) Substitute \(x = \frac{2}{3}\):
\(p\left(\frac{2}{3}\right) = 3\left(\frac{2}{3}\right) - 2 = 2 - 2 = 0\) (Yes)
(b) Substitute \(x = 3\) and \(x = -3\):
\(p(3) = (3)^2 - 9 = 9 - 9 = 0\)
\(p(-3) = (-3)^2 - 9 = 9 - 9 = 0\) (Yes)
(c) Substitute \(x = \pi\):
\(p(\pi) = 5(\pi) - \pi = 4\pi \neq 0\) (No)
In simple words: Evaluate the polynomial at the given value; if the result is 0, it is a zero.

Exam Tip: Do not panic seeing constants like \(\pi\); treat them like regular numbers during substitution.

Q8. Find the Zeroes of the polynomial in each of the following
(a) \(cx + d\)
(b) \(x^2 + 5x + 6\)
(c) \((x + 2)(x + 3)\)
Answer: (a) \(-\frac{d}{c}\), (b) \(-3, -2\), (c) \(-2, -3\)
(a) Set \(cx + d = 0 \Rightarrow cx = -d \Rightarrow x = -\frac{d}{c}\).
(b) Set \(x^2 + 5x + 6 = 0\)
\(x^2 + 3x + 2x + 6 = 0\)
\(x(x + 3) + 2(x + 3) = 0\)
\((x + 3)(x + 2) = 0\)
So, \(x = -3\) or \(x = -2\).
(c) Set \((x + 2)(x + 3) = 0\)
\(x + 2 = 0 \Rightarrow x = -2\) or \(x + 3 = 0 \Rightarrow x = -3\).
In simple words: To find zeroes, equate the polynomial expression to zero and solve for \(x\).

Exam Tip: For quadratic polynomials, factorisation by splitting the middle term is the standard method.

Q9. How many zeroes a linear polynomial have?
Answer: 1
A linear polynomial is of the form \(ax + b\) (where \(a \neq 0\)), which has a degree of \(1\). Therefore, it has exactly one zero.
In simple words: The number of zeroes of a polynomial is equal to its degree.

Q10. Can a polynomial have more than one zeroes?
Answer: Yes
A polynomial of degree \(n\) (where \(n > 1\)), such as a quadratic or cubic polynomial, can have more than one zero. For example, a quadratic polynomial has 2 zeroes.
In simple words: Yes, except for linear polynomials (which have exactly one), polynomials of higher degrees can have multiple zeroes.

Exam Tip: Quadratic polynomials have up to 2 zeroes, and cubic polynomials have up to 3 zeroes.