NCERT Solutions for Class 9 Maths Chapter 2 Exercise 2.3 - Polynomials

Q1 (i) Determine which of the following polynomials has $(x + 1)$ a factor : $x^3 + x^2 +x + 1$

Answer:

Let p(x) = x³ + x² + x +1

Use the Factor Theorem, which says:

If p(–1) = 0, then (x + 1) is a factor of the polynomial p(x).

So, p(−1) = (−1)³ + (−1)² + (−1) + 1 = -1 + 1 - 1 + 1 = 0

Therefore, Yes, (x + 1) is a factor of polynomial $p(x)=x^3 + x^2 +x + 1$

Q1 (ii) Determine which of the following polynomials has $(x + 1)$ a factor : $x^4 + x^3 + x^2 +x + 1$

Answer:

Let p(x) = x⁴ + x³ + x² + x + 1

Use the Factor Theorem, which says:

If p(–1) = 0, then (x + 1) is a factor of the polynomial p(x).

So, p(−1) = (−1)⁴ + (−1)³ + (−1)² + (−1) + 1 = 1 - 1 + 1 - 1 + 1 = 1

Therefore, No, (x + 1) is not a factor of polynomial $p(x)=x^4 + x^3 + x^2 +x + 1$

Q1 (iii) Determine which of the following polynomials has $(x + 1)$ a factor : $x^4 + 3x^3 + 3x^2 +x + 1$

Answer:

Let p(x)= x⁴ + 3x³ + 3x² + x + 1

Use the Factor Theorem, which says:

If p(–1) = 0, then (x + 1) is a factor of the polynomial p(x).

So, p(−1) = (−1)⁴ + 3(−1)³ + 3(−1)² + (−1) + 1 = 1 - 3 + 3 - 1 + 1 = 1

Therefore, No, (x + 1) is not a factor of polynomial $p(x)=x^4 + 3x^3 + 3x^2 +x + 1$

Q1 (iv) Determine which of the following polynomials has $(x + 1)$ a factor : $x^3 - x^2 -(2 + \sqrt2)x + \sqrt2$

Answer:

Let p(x) = $x^3 - x^2 -(2 + \sqrt2)x + \sqrt2$

Use the Factor Theorem, which says:

If p(–1) = 0, then (x + 1) is a factor of the polynomial p(x).

So, $p(-1) = (-1)^3 - (-1)^2 - (2+\sqrt{2})(-1) + \sqrt{2}$ = $-1 - 1 + 2 + \sqrt{2} + \sqrt{2}$ = $2\sqrt{2} \neq 0$

Therefore, No, (x + 1) is not a factor of polynomial p(x) = $x^3 - x^2 -(2 + \sqrt2)x + \sqrt2$

Q2 (i) Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case: $p(x) = 2x^3 + x^2 - 2x - 1,\ g(x) = x + 1$

Answer:

g(x) = 0

⇒ x + 1 = 0

⇒ x = −1

Now, p(−1) = 2(−1)³ + (−1)² – 2(−1) –1 = −2 + 1 + 2 −1 = 0

Therefore, Yes, g(x) is a factor of p(x)

Q2 (ii) Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case: $p(x) = x^3 + 3x^2 + 3x + 1, \ g(x) = x + 2$

Answer:

g(x) = 0

⇒ x + 2 = 0

⇒ x = −2

Now,

p(−2) = (−2)³ + 3(−2)² + 3(−2) + 1

= − 8 + 12 − 6 + 1

= −1 ≠ 0

Therefore, No, g(x) is not a factor of p(x)

Q2 (iii) Use the Factor Theorem to determine whether g(x) is a factor of p(x) in the following case: $p(x) = x^3 - 4x^2 + x + 6, \ g(x) = x - 3$

Answer:

g(x) = 0

⇒ x − 3 = 0

⇒ x = 3

Now,

p(3) = (3)³ − 4(3)² + (3) + 6

= 27 − 36 + 3 + 6

= 0

Therefore, yes, g(x) is a factor of p(x).

Q3 (i) Find the value of k , if $x - 1$ is a factor of p(x) in the following case: $p(x) = x^2 + x + k$

Answer:

As (x - 1) is a factor of p(x), it means that p(1) = 0

So, By the Factor Theorem

⇒ (1)² + (1) + k = 0

⇒ 1 + 1 + k = 0

⇒ 2 + k = 0

⇒ k = −2

Q3 (ii) Find the value of k , if $x - 1$ is a factor of p(x) in the following case: $p(x) = 2x^2 + kx + \sqrt2$

Answer:

As (x - 1) is a factor of p(x), it means that p(1) = 0

So, by the factor theorem

⇒ $2(1)^2 + k(1) + \sqrt2$ = 0

⇒ $2 + k + \sqrt2$ = 0

⇒ k = $-2 - \sqrt2$

Q3 (iii) Find the value of k , if $x - 1$ is a factor of p(x) in the following case: $p(x) = kx^2-\sqrt2 x + 1$

Answer:

As (x - 1) is a factor of p(x), it means that p(1) = 0

So, by the factor theorem

⇒ $k(1)^2-\sqrt2 (1) + 1$ = 0

⇒ k = $ - 1 + \sqrt2$

Q3 (iv) the value of k , if $x - 1$ is a factor of p(x) in the following case: $p(x) = kx^2 -3 x + k$

Answer:

As (x - 1) is a factor of p(x), it means that p(1) = 0

So, by the factor theorem

⇒ $k(1)^2 -3 (1) + k$ = 0

⇒ k − 3 + k = 0

⇒ 2k − 3 = 0

⇒ k = $\frac{3}{2}$

Q4 (i) Factorise : $12x^2 - 7x + 1$

Answer:

Given polynomial is $12x^2 - 7x + 1$

We need to factorise the middle term into two terms such that their product is equal to $12 \times 1 = 12$ and their sum is equal to $-7$

We can solve it as

$\Rightarrow 12x^2 - 7x + 1$

$\Rightarrow 12x^2-3x-4x+1$ $(\because -3 \times -4 = 12 \ \ and \ \ -3+(-4) = -7)$

$\Rightarrow 3x(4x-1)-1(4x-1)$

$\Rightarrow (3x-1)(4x-1)$

Q4 (ii) Factorise : $2x^2 + 7x + 3$

Answer:

Given polynomial is $2x^2 + 7x + 3$

We need to factorise the middle term into two terms such that their product is equal to $2 \times 3 = 6$ and their sum is equal to $7$

We can solve it as

$\Rightarrow 12x^2 - 7x + 1$

$\Rightarrow 2x^2+6x+x+3$ $(\because 6 \times 1 = 6 \ \ and \ \ 6+1 = 7)$

$\Rightarrow 2x(x+3)+1(x+3)$

$\Rightarrow (2x+1)(x+3)$

Q4 (iii) Factorise : $6x^2 +5x - 6$

Answer:

Given polynomial is $6x^2 +5x - 6$

We need to factorise the middle term into two terms such that their product is equal to $6 \times -6 =-3 6$ and their sum is equal to $5$

We can solve it as

$\Rightarrow 6x^2 +5x - 6$

$\Rightarrow 6x^2 +9x -4x - 6$ $(\because 9 \times -4 = -36 \ \ and \ \ 9+(-4) = 5)$

$\Rightarrow 3x(2x+3)-2(2x+3)$

$\Rightarrow (2x+3)(3x-2)$

Q4 (iv) Factorise : $3x^2 - x - 4$

Answer:

Given polynomial is $3x^2 - x - 4$

We need to factorise the middle term into two terms such that their product is equal to $3 \times -4 =-12$ and their sum is equal to $-1$

We can solve it as

$\Rightarrow 3x^2 - x - 4$

$\Rightarrow 3x^2 -4 x+3x - 4$ $(\because 3 \times -4 = -12 \ \ and \ \ 3+(-4) = -1)$

$\Rightarrow x(3x-4)+1(3x-4)$

$\Rightarrow (x+1)(3x-4)$

Q5 (i) Factorise : $x^3 - 2x^2 - x +2$

Answer:

Given polynomial is $x^3 - 2x^2 - x +2$

Now, by the hit and trial method, we observed that $(x+1)$ is one of the factors of the given polynomial.

By the long division method, we will get

We know that Dividend = (Divisor $\times$ Quotient) + Remainder

$x^3 - 2x^2 - x +2 = (x+1)(x^2-3x+2)+0$

$= (x+1)(x^2-2x-x+2)$

$= (x+1)(x-2)(x-1)$

Therefore, on factorization of $x^3 - 2x^2 - x +2$ we will get $(x+1)(x-2)(x-1)$

Q5 (ii) Factorise : $x^3 - 3x^2 -9x -5$

Answer:

Given polynomial is $x^3 - 3x^2 -9x -5$

Now, by the hit and trial method, we observed that $(x+1)$ is one of the factors of the given polynomial.

By the long division method, we will get

We know that Dividend = (Divisor $\times$ Quotient) + Remainder

$x^3 - 3x^2 -9x -5=(x+1)(x^2-4x-5)$

$= (x+1)(x^2-5x+x-5)$

$= (x+1)(x-5)(x+1)$

Therefore, on factorization of $x^3 - 3x^2 -9x -5$ we will get $(x+1)(x-5)(x+1)$

Q5 (iii) Factorise : $x^3 + 13x^2 + 32x + 20$

Answer:

Given polynomial is $x^3 + 13x^2 + 32x + 20$

Now, by the hit and trial method, we observed that $(x+1)$ is one of the factors of the given polynomial.

By the long division method, we will get

We know that Dividend = (Divisor $\times$ Quotient) + Remainder

$x^3 + 13x^2 + 32x + 20=(x+1)(x^2+12x+20)$

$= (x+1)(x^2+10x+2x+20)$

$= (x+1)(x+10)(x+2)$

Therefore, on factorization of $x^3 + 13x^2 + 32x + 20$ we will get $(x+1)(x+10)(x+2)$

Q5 (iv) Factorise : $2y^3 + y^2 - 2y - 1$

Answer:

Given polynomial is $2y^3 + y^2 - 2y - 1$

Now, by the hit and trial method, we observed that $(y-1)$ is one of the factors of the given polynomial.

By the long division method, we will get

We know that Dividend = (Divisor $\times$ Quotient) + Remainder

$2y^3 + y^2 - 2y - 1= (y-1)(2y^2+3y+1)$

$= (y-1)(2y^2+2y+y+1)$

$= (y-1)(2y+1)(y+1)$

Therefore, on factorization of $2y^3 + y^2 - 2y - 1$ we will get $(y-1)(2y+1)(y+1)$