RBSE Solutions Class 9 Maths Chapter 3 Polynomial Exercise 3.4

Polynomial Ex 3.4

Question 1. Determine which of the following (RBSESolutions.com) polynomials has x – 1 factor.
(i) \( x^4 - 2x^3 - 3x^2 + 2x + 2 \)
(ii) \( x^4 + x^3 + x^2 + x + 1 \)
(iii) \( x^4 + 3x^3 - 3x^2 + x - 2 \)
(iv) \( x^3 – x^2 - (2 + \sqrt{3})x + \sqrt{3} \)
Answer:
According to the Factor Theorem, if \( (x - a) \) is a factor of a polynomial \( p(x) \), then \( p(a) = 0 \). Here, we are checking if \( (x - 1) \) is a factor, so we need to evaluate the polynomial at \( x = 1 \). If \( p(1) = 0 \), then \( (x - 1) \) is a factor.

(i) Let \( p(x) = x^4 – 2x^3 - 3x^2 + 2x + 2 \)
Now, we find \( p(1) \):
\( p(1) = (1)^4 – 2(1)^3 – 3(1)^2 + 2(1) + 2 \)
\( = 1 - 2 - 3 + 2 + 2 \)
\( = 0 \)
Since \( p(1) = 0 \), therefore \( (x - 1) \) is a factor of \( x^4 - 2x^3 - 3x^2 + 2x + 2 \).

(ii) Let \( p(x) = x^4 + x^3 + x^2 + x + 1 \)
Now, we find \( p(1) \):
\( p(1) = (1)^4 + (1)^3 + (1)^2 + (1) + 1 \)
\( = 1 + 1 + 1 + 1 + 1 \)
\( = 5 \)
Since \( p(1) \ne 0 \), therefore \( (x - 1) \) is not a factor of \( x^4 + x^3 + x^2 + x + 1 \).

(iii) Let \( p(x) = x^4 + 3x^3 - 3x^2 + x - 2 \)
Now, we find \( p(1) \):
\( p(1) = (1)^4 + 3(1)^3 – 3(1)^2 + (1) - 2 \)
\( = 1 + 3 - 3 + 1 - 2 \)
\( = 0 \)
Since \( p(1) = 0 \), therefore \( (x - 1) \) is a factor of \( x^4 + 3x^3 - 3x^2 + x - 2 \).

(iv) Let \( p(x) = x^3 – x^2 - (2 + \sqrt{3})x + \sqrt{3} \)
Now, we find \( p(1) \):
\( p(1) = (1)^3 – (1)^2 - (2 + \sqrt{3})(1) + \sqrt{3} \)
\( = 1 - 1 - (2 + \sqrt{3}) + \sqrt{3} \)
\( = 0 - 2 - \sqrt{3} + \sqrt{3} \)
\( = -2 \)
Since \( p(1) \ne 0 \), therefore \( (x - 1) \) is not a factor of \( x^3 – x^2 - (2 + \sqrt{3})x + \sqrt{3} \).
In simple words: To find out if \( (x-1) \) is a factor, we put \( x=1 \) into each polynomial. If the answer is zero, then \( (x-1) \) is a factor. We found that for polynomials (i) and (iii), putting \( x=1 \) gave zero, so \( (x-1) \) is their factor.

🎯 Exam Tip: Remember the Factor Theorem: \( (x-a) \) is a factor of \( p(x) \) if and only if \( p(a) = 0 \). This is a fundamental concept for checking factors of polynomials.

Question 2. Use the factor theorem (RBSESolutions.com) to determine whether g(x) is a factor of p(x) in each of the following case.
(i) \( p(x) = 3x^3 - x^2 – 3x + 1; g(x) = x + 1 \)
(ii) \( p(x) = 2x^4 – 7x^3 – 13x^2 + 63x – 45; g(x) = x − 1 \)
(iii) \( p(x) = 3x^3 + 3x^2 + 3x + 1; g(x) = x + 2 \)
(iv) \( p(x) = 2x^3 + x^2 – 2x – 1; g(x) = 2x + 1 \)
Answer:
We will use the Factor Theorem. For \( g(x) \) to be a factor of \( p(x) \), \( p(a) \) must be \( 0 \), where \( a \) is the root of \( g(x) \). This means if \( g(x) = x - a \), we check \( p(a) \). If \( g(x) = x + a \), we check \( p(-a) \). If \( g(x) = bx + a \), we check \( p(-\frac{a}{b}) \).

(i) Given \( p(x) = 3x^3 - x^2 – 3x + 1 \) and \( g(x) = x + 1 \).
Set \( g(x) = 0 \Rightarrow x + 1 = 0 \Rightarrow x = -1 \).
Now, calculate \( p(-1) \):
\( p(-1) = 3(-1)^3 – (-1)^2 – 3(-1) + 1 \)
\( = 3(-1) - (1) + 3 + 1 \)
\( = -3 - 1 + 3 + 1 \)
\( = 0 \)
Since \( p(-1) = 0 \), therefore \( g(x) = x + 1 \) is a factor of \( p(x) \).

(ii) Given \( p(x) = 2x^4 – 7x^3 – 13x^2 + 63x – 45 \) and \( g(x) = x - 1 \).
Set \( g(x) = 0 \Rightarrow x - 1 = 0 \Rightarrow x = 1 \).
Now, calculate \( p(1) \):
\( p(1) = 2(1)^4 – 7(1)^3 – 13(1)^2 + 63(1) – 45 \)
\( = 2 - 7 - 13 + 63 - 45 \)
\( = -65 + 65 \)
\( = 0 \)
Since \( p(1) = 0 \), therefore \( g(x) = x - 1 \) is a factor of \( p(x) \).

(iii) Given \( p(x) = 3x^3 + 3x^2 + 3x + 1 \) and \( g(x) = x + 2 \).
Set \( g(x) = 0 \Rightarrow x + 2 = 0 \Rightarrow x = -2 \).
Now, calculate \( p(-2) \):
\( p(-2) = 3(-2)^3 + 3(-2)^2 + 3(-2) + 1 \)
\( = 3(-8) + 3(4) - 6 + 1 \)
\( = -24 + 12 - 6 + 1 \)
\( = -30 + 13 \)
\( = -17 \)
Since \( p(-2) \ne 0 \), therefore \( g(x) = x + 2 \) is not a factor of \( p(x) \).

(iv) Given \( p(x) = 2x^3 + x^2 – 2x – 1 \) and \( g(x) = 2x + 1 \).
Set \( g(x) = 0 \Rightarrow 2x + 1 = 0 \Rightarrow x = -\frac{1}{2} \).
Now, calculate \( p(-\frac{1}{2}) \):
\( p(-\frac{1}{2}) = 2(-\frac{1}{2})^3 + (-\frac{1}{2})^2 – 2(-\frac{1}{2}) – 1 \)
\( = 2(-\frac{1}{8}) + (\frac{1}{4}) - (-1) - 1 \)
\( = -\frac{1}{4} + \frac{1}{4} + 1 - 1 \)
\( = 0 \)
Since \( p(-\frac{1}{2}) = 0 \), therefore \( g(x) = 2x + 1 \) is a factor of \( p(x) \).
In simple words: To see if \( g(x) \) is a factor of \( p(x) \), we find the value of \( x \) that makes \( g(x) \) zero. Then we put this \( x \) value into \( p(x) \). If \( p(x) \) also becomes zero, then \( g(x) \) is a factor. We tested each pair and found that \( g(x) \) is a factor for (i), (ii), and (iv), but not for (iii).

🎯 Exam Tip: When \( g(x) \) is in the form \( ax+b \), remember to find the root \( x = -\frac{b}{a} \) before substituting into \( p(x) \).

Question 3. Find the (RBSESolutions.com) value of k, when x – 5 is a factor of \( x^3 – 3x^2 + kx – 10 \).
Answer:
Let the polynomial be \( p(x) = x^3 - 3x^2 + kx - 10 \).
Since \( (x - 5) \) is a factor of \( p(x) \), according to the Factor Theorem, \( p(5) \) must be equal to \( 0 \).
So, we substitute \( x = 5 \) into the polynomial and set it to zero:
\( p(5) = (5)^3 – 3(5)^2 + k(5) - 10 = 0 \)
Now, we solve for \( k \):
\( 125 - 3(25) + 5k - 10 = 0 \)
\( 125 - 75 + 5k - 10 = 0 \)
\( 50 + 5k - 10 = 0 \)
\( 40 + 5k = 0 \)
\( 5k = -40 \)
\( k = \frac{-40}{5} \)
\( k = -8 \)
Thus, the value of \( k \) is \( -8 \). Knowing the factor helps us find unknown coefficients.
In simple words: Because \( (x-5) \) is a factor, we know that if we put \( x=5 \) into the polynomial, the answer should be zero. We solve this equation to find the number \( k \). We found \( k \) is \( -8 \).

🎯 Exam Tip: Always set the polynomial equal to zero when a factor is given, then solve the resulting equation for the unknown constant.

Question 4. Find the value of k, when x − 1 is a factor of \( 2x^2 + kx + \sqrt{2} \).
Answer:
Let the polynomial be \( p(x) = 2x^2 + kx + \sqrt{2} \).
Since \( (x - 1) \) is a factor of \( p(x) \), by the Factor Theorem, \( p(1) \) must be \( 0 \).
Substitute \( x = 1 \) into the polynomial and set it equal to zero:
\( p(1) = 2(1)^2 + k(1) + \sqrt{2} = 0 \)
Now, solve for \( k \):
\( 2(1) + k + \sqrt{2} = 0 \)
\( 2 + k + \sqrt{2} = 0 \)
\( k = -2 - \sqrt{2} \)
\( k = -(2 + \sqrt{2}) \)
Therefore, the value of \( k \) is \( -(2 + \sqrt{2}) \). This method is widely used for determining unknown coefficients in polynomials.
In simple words: Since \( (x-1) \) is a factor, when we put \( x=1 \) into the polynomial, the answer must be zero. We then solve the equation to find \( k \). We get \( k = -(2 + \sqrt{2}) \).

🎯 Exam Tip: Be careful with signs when substituting negative values or dealing with terms like \( -\sqrt{2} \). Double-check your arithmetic, especially with square roots.

Question 5. If \( (x + 1) \) and \( (x - 1) \) are the (RBSESolutions.com) factors of \( x^4 + ax^3 – 3x^2 + 2x + b \) then find the values of a and b.
Answer:
Let the polynomial be \( p(x) = x^4 + ax^3 – 3x^2 + 2x + b \).

Given that \( (x + 1) \) is a factor of \( p(x) \). By the Factor Theorem, \( p(-1) = 0 \).
Substitute \( x = -1 \) into \( p(x) \):
\( (-1)^4 + a(-1)^3 – 3(-1)^2 + 2(-1) + b = 0 \)
\( 1 - a - 3 - 2 + b = 0 \)
\( -a + b - 4 = 0 \)
\( b - a = 4 \) ...(i)

Given that \( (x - 1) \) is also a factor of \( p(x) \). By the Factor Theorem, \( p(1) = 0 \).
Substitute \( x = 1 \) into \( p(x) \):
\( (1)^4 + a(1)^3 – 3(1)^2 + 2(1) + b = 0 \)
\( 1 + a - 3 + 2 + b = 0 \)
\( a + b = 0 \) ...(ii)

Now we have a system of two linear equations with two variables, \( a \) and \( b \). We can solve them simultaneously.
Equation (i): \( -a + b = 4 \)
Equation (ii): \( a + b = 0 \)

Add Equation (i) and Equation (ii):
\( (-a + b) + (a + b) = 4 + 0 \)
\( 2b = 4 \)
\( b = \frac{4}{2} \)
\( b = 2 \)

Substitute the value of \( b = 2 \) into Equation (ii):
\( a + 2 = 0 \)
\( a = -2 \)

Therefore, the values are \( a = -2 \) and \( b = 2 \). This method helps find multiple unknown coefficients.
In simple words: We know that if \( (x+1) \) and \( (x-1) \) are factors, then putting \( x=-1 \) and \( x=1 \) into the polynomial must give zero. We do this to get two simple equations with \( a \) and \( b \). Then we solve these two equations together to find that \( a \) is \( -2 \) and \( b \) is \( 2 \).

🎯 Exam Tip: When given multiple factors, you will usually get a system of equations. Solve these equations carefully to find all unknown constants.

Question 6. Factorise by splitting (RBSESolutions.com) middle term
(i) \( 3x^2 + 7x + 2 \)
(ii) \( 4x^2 - x - 3 \)
(iii) \( 12x^2 - 7x + 1 \)
(iv) \( 6x^2 + 5x - 6 \)
Answer:
To factorise quadratic polynomials by splitting the middle term \( bx \), we need to find two numbers that multiply to \( ac \) (the product of the first and last coefficients) and add up to \( b \) (the middle coefficient).

(i) Let \( p(x) = 3x^2 + 7x + 2 \)
Here, \( a = 3, b = 7, c = 2 \). We need two numbers whose product is \( ac = 3 \times 2 = 6 \) and whose sum is \( b = 7 \). The numbers are \( 6 \) and \( 1 \).
Split the middle term \( 7x \) as \( 6x + x \):
\( 3x^2 + 7x + 2 = 3x^2 + 6x + x + 2 \)
Now, group the terms and factor out common factors:
\( = 3x(x + 2) + 1(x + 2) \)
\( = (x + 2)(3x + 1) \)

(ii) Let \( p(x) = 4x^2 - x - 3 \)
Here, \( a = 4, b = -1, c = -3 \). We need two numbers whose product is \( ac = 4 \times (-3) = -12 \) and whose sum is \( b = -1 \). The numbers are \( -4 \) and \( 3 \).
Split the middle term \( -x \) as \( -4x + 3x \):
\( 4x^2 - x - 3 = 4x^2 - 4x + 3x - 3 \)
Now, group the terms and factor out common factors:
\( = 4x(x - 1) + 3(x - 1) \)
\( = (x - 1)(4x + 3) \)

(iii) Let \( p(x) = 12x^2 - 7x + 1 \)
Here, \( a = 12, b = -7, c = 1 \). We need two numbers whose product is \( ac = 12 \times 1 = 12 \) and whose sum is \( b = -7 \). The numbers are \( -4 \) and \( -3 \).
Split the middle term \( -7x \) as \( -4x - 3x \):
\( 12x^2 - 7x + 1 = 12x^2 - 4x - 3x + 1 \)
Now, group the terms and factor out common factors:
\( = 4x(3x - 1) - 1(3x - 1) \)
\( = (3x - 1)(4x - 1) \)

(iv) Let \( p(x) = 6x^2 + 5x - 6 \)
Here, \( a = 6, b = 5, c = -6 \). We need two numbers whose product is \( ac = 6 \times (-6) = -36 \) and whose sum is \( b = 5 \). The numbers are \( 9 \) and \( -4 \).
Split the middle term \( 5x \) as \( 9x - 4x \):
\( 6x^2 + 5x - 6 = 6x^2 + 9x - 4x - 6 \)
Now, group the terms and factor out common factors:
\( = 3x(2x + 3) - 2(2x + 3) \)
\( = (2x + 3)(3x - 2) \)
In simple words: To factor these expressions, we look for two numbers. These numbers must multiply to give the first number times the last number, and they must add up to the middle number. Once we find these numbers, we split the middle part of the expression and group terms to find the factors.

🎯 Exam Tip: Always double-check your factors by multiplying them back out to ensure you get the original polynomial. Pay close attention to signs when finding the two numbers.

Question 7. Find the zeroes of (RBSESolutions.com) the polynomials:
(i) \( x^3 + 6x^2 + 11x + 6 \)
(ii) \( x^3 + 2x^2 − x − 2 \)
(iii) \( x^4 - 2x^3 - 7x^2 + 8x + 12 \)
(iv) \( x^3 – 2x^3 − x +2 \)
(v) \( x^3-3x^2 - 9x - 5 \)
(vi) \( x^3 – 23x^2 + 142x - 120 \)
Answer:
To find the zeroes of a polynomial, we look for values of \( x \) that make the polynomial equal to zero. For polynomials with integer coefficients and a leading coefficient of 1, any rational zeroes must be factors of the constant term. We can test these factors using the Factor Theorem.

(i) Let \( p(x) = x^3 + 6x^2 + 11x + 6 \).
The constant term is \( 6 \). Its factors are \( \pm 1, \pm 2, \pm 3, \pm 6 \).
Try \( x = -1 \):
\( p(-1) = (-1)^3 + 6(-1)^2 + 11(-1) + 6 \)
\( = -1 + 6(1) - 11 + 6 \)
\( = -1 + 6 - 11 + 6 = 0 \)
So, \( -1 \) is a zero of \( p(x) \).

Try \( x = -2 \):
\( p(-2) = (-2)^3 + 6(-2)^2 + 11(-2) + 6 \)
\( = -8 + 6(4) - 22 + 6 \)
\( = -8 + 24 - 22 + 6 = 0 \)
So, \( -2 \) is a zero of \( p(x) \).

Try \( x = -3 \):
\( p(-3) = (-3)^3 + 6(-3)^2 + 11(-3) + 6 \)
\( = -27 + 6(9) - 33 + 6 \)
\( = -27 + 54 - 33 + 6 = 0 \)
So, \( -3 \) is a zero of \( p(x) \).
The integral zeroes of the polynomial are \( -1, -2, -3 \).

(ii) Let \( p(x) = x^3 + 2x^2 − x − 2 \).
The constant term is \( -2 \). Its factors are \( \pm 1, \pm 2 \).
Try \( x = 1 \):
\( p(1) = (1)^3 + 2(1)^2 - (1) - 2 \)
\( = 1 + 2(1) - 1 - 2 \)
\( = 1 + 2 - 1 - 2 = 0 \)
So, \( 1 \) is a zero of \( p(x) \).

Try \( x = -1 \):
\( p(-1) = (-1)^3 + 2(-1)^2 - (-1) - 2 \)
\( = -1 + 2(1) + 1 - 2 \)
\( = -1 + 2 + 1 - 2 = 0 \)
So, \( -1 \) is a zero of \( p(x) \).

Try \( x = -2 \):
\( p(-2) = (-2)^3 + 2(-2)^2 - (-2) - 2 \)
\( = -8 + 2(4) + 2 - 2 \)
\( = -8 + 8 + 2 - 2 = 0 \)
So, \( -2 \) is a zero of \( p(x) \).
The integral zeroes of the polynomial are \( 1, -1, -2 \).

(iii) Let \( p(x) = x^4 - 2x^3 - 7x^2 + 8x + 12 \).
The constant term is \( 12 \). Its factors are \( \pm 1, \pm 2, \pm 3, \pm 4, \pm 6, \pm 12 \).
Try \( x = -1 \):
\( p(-1) = (-1)^4 - 2(-1)^3 - 7(-1)^2 + 8(-1) + 12 \)
\( = 1 - 2(-1) - 7(1) - 8 + 12 \)
\( = 1 + 2 - 7 - 8 + 12 = 0 \)
So, \( -1 \) is a zero of \( p(x) \).

Try \( x = -2 \):
\( p(-2) = (-2)^4 - 2(-2)^3 - 7(-2)^2 + 8(-2) + 12 \)
\( = 16 - 2(-8) - 7(4) - 16 + 12 \)
\( = 16 + 16 - 28 - 16 + 12 = 0 \)
So, \( -2 \) is a zero of \( p(x) \).

Try \( x = 2 \):
\( p(2) = (2)^4 - 2(2)^3 - 7(2)^2 + 8(2) + 12 \)
\( = 16 - 2(8) - 7(4) + 16 + 12 \)
\( = 16 - 16 - 28 + 16 + 12 = 0 \)
So, \( 2 \) is a zero of \( p(x) \).

Try \( x = 3 \):
\( p(3) = (3)^4 - 2(3)^3 - 7(3)^2 + 8(3) + 12 \)
\( = 81 - 2(27) - 7(9) + 24 + 12 \)
\( = 81 - 54 - 63 + 24 + 12 = 0 \)
So, \( 3 \) is a zero of \( p(x) \).
The integral zeroes of the polynomial are \( -1, -2, 2, 3 \).

(iv) Let \( p(x) = x^3 – 2x^3 − x + 2 = -x^3 - x + 2 \).
The constant term is \( 2 \). Its factors are \( \pm 1, \pm 2 \).
Try \( x = 1 \):
\( p(1) = -(1)^3 - (1) + 2 \)
\( = -1 - 1 + 2 = 0 \)
So, \( 1 \) is a zero of \( p(x) \).

Try \( x = -1 \):
\( p(-1) = -(-1)^3 - (-1) + 2 \)
\( = -(-1) + 1 + 2 \)
\( = 1 + 1 + 2 = 4 \)
So, \( -1 \) is not a zero of \( p(x) \).

Try \( x = 2 \):
\( p(2) = -(2)^3 - (2) + 2 \)
\( = -8 - 2 + 2 = -8 \)
So, \( 2 \) is not a zero of \( p(x) \).

Try \( x = -2 \):
\( p(-2) = -(-2)^3 - (-2) + 2 \)
\( = -(-8) + 2 + 2 \)
\( = 8 + 2 + 2 = 12 \)
So, \( -2 \) is not a zero of \( p(x) \).
The only integral zero of the polynomial is \( 1 \).

(v) Let \( p(x) = x^3-3x^2 - 9x - 5 \).
The constant term is \( -5 \). Its factors are \( \pm 1, \pm 5 \).
Try \( x = -1 \):
\( p(-1) = (-1)^3 - 3(-1)^2 - 9(-1) - 5 \)
\( = -1 - 3(1) + 9 - 5 \)
\( = -1 - 3 + 9 - 5 = 0 \)
So, \( -1 \) is a zero of \( p(x) \).

Try \( x = 5 \):
\( p(5) = (5)^3 - 3(5)^2 - 9(5) - 5 \)
\( = 125 - 3(25) - 45 - 5 \)
\( = 125 - 75 - 45 - 5 = 0 \)
So, \( 5 \) is a zero of \( p(x) \).
The integral zeroes of the polynomial are \( -1, 5 \).

(vi) Let \( p(x) = x^3 – 23x^2 + 142x - 120 \).
The constant term is \( -120 \). Its factors are many, but we will test common ones.
Try \( x = 1 \):
\( p(1) = (1)^3 - 23(1)^2 + 142(1) - 120 \)
\( = 1 - 23 + 142 - 120 \)
\( = 143 - 143 = 0 \)
So, \( 1 \) is a zero of \( p(x) \).

Try \( x = 10 \):
\( p(10) = (10)^3 - 23(10)^2 + 142(10) - 120 \)
\( = 1000 - 23(100) + 1420 - 120 \)
\( = 1000 - 2300 + 1420 - 120 \)
\( = 2420 - 2420 = 0 \)
So, \( 10 \) is a zero of \( p(x) \).

Try \( x = 12 \):
\( p(12) = (12)^3 - 23(12)^2 + 142(12) - 120 \)
\( = 1728 - 23(144) + 1704 - 120 \)
\( = 1728 - 3312 + 1704 - 120 \)
\( = 3432 - 3432 = 0 \)
So, \( 12 \) is a zero of \( p(x) \).
The integral zeroes of the polynomial are \( 1, 10, 12 \).
In simple words: To find the zeroes, we look for numbers that make the polynomial equal to zero. For polynomials where the first term has no number in front (or just a 1), we check the numbers that can divide the last term evenly. We plug in these numbers one by one until we find the ones that give zero as a result. These numbers are the zeroes.

🎯 Exam Tip: For higher-degree polynomials (degree 3 or more), finding one zero through trial and error often allows you to divide the polynomial by \( (x-a) \) (where \( a \) is the zero) to get a simpler quadratic, which can then be factorised further.