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# Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:

(i) p(x) = 2x^{3} + x^{2} - 2x - 1, g(x) = x +1

(ii) p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

(iii) p(x) = x^{3} - 4x^{2} + x + 6, g(x) = x - 3

**Solution:**

According to factor theorem, (x - a) is a factor of a polynomial p(x) if p(a) = 0.

To find if g(x ) = x + a is a factor of p(x), we need to find the root of g(x).

x + a = 0

⇒ x = -a

**(i)** Let p(x) = 2x^{3} + x^{2} - 2x - 1, g(x) = x + 1

x + 1 = 0

⇒ x = -1

Now,

p(-1) = 2(-1)^{3} + (-1)^{2} - 2(-1) - 1

= -2 + 1 + 2 - 1

= 0

Since the remainder of p(-1) = 0 , by factor theorem we can say that g(x) = x + 1 is a factor of p(x) = 2x^{3} + x^{2} - 2x - 1.

**(ii)** Let p(x) = x^{3} + 3x^{2} + 3x + 1, g(x) = x + 2

x + 2 = 0

⇒ x = -2

Now,

p(-2) = (-2)^{3} + 3(-2)^{2} + 3(-2) + 1

= -8 + 12 - 6 + 1

= -1 ≠ 0

Since the remainder of p(-2) ≠ 0 , by factor theorem we can say g(x) = x + 2 is not a factor of p(x) = x^{3} + 3x^{2} + 3x + 1.

**(iii)** Let p(x) = x^{3} - 4x^{2} + x + 6, g(x) = x - 3

x - 3 = 0

⇒ x = 3

Now,

p(3) = (3)^{3} - 4(3)^{2} + 3 + 6

= 27 - 36 + 3 + 6

= 0

Since the remainder of p(3) = 0 , by factor theorem we can say g(x) = x - 3 is a factor of p(x) = x^{3} - 4x^{2} + x + 6

**☛ Check: **NCERT Solutions Class 9 Maths Chapter 2

**Video Solution:**

## Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases: (i) p(x) = 2x³ + x² - 2x - 1, g(x) = x + 1 (ii) p(x) = x³ + 3x² + 3x + 1, g(x) = x + 2 (iii) p(x) = x³ - 4x² + x + 6, g(x) = x - 3

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.4 Question 2

**Summary:**

Using factor theorem we see that, g(x) = x + 1 is a factor of p(x) = 2x^{3} + x^{2} - 2x - 1, g(x) = x + 2 is not a factor of p(x) = x^{3} + 3x^{2} + 3x + 1, and g(x) = x - 3 is a factor of p(x) = x^{3} - 4x^{2} + x + 6.

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