# Polynomials

## Exercise 2.4 Part 2

Question 2: 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

Answer: Let g(x) = 0
Or, x + 1 = 0
Or, x = - 1

This means -1 is the zero of the given polynomial g(x) = x + 1
Now, according to Factor Theorem, if p(-1) is equal to zero, then g(x) is the factor of p(x) = 2x^3 + x^2 – 2x – 1
Or, p( - 1) = 2( - 1)^3 + ( - 1)^2 – 2( - 1) – 1
= 2 x ( - 1) + 1 + 2 – 1
= - 2 + 1 + 2 – 1 = 0
i.e. p( - 1) = 0

Since, p( - 1) = 0, therefore, g(x) = (x + 1) is a factor of p(x) = 2x^3 + x^2 – 2x - 1

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

Answer: Let g(x) = 0
Or, x + 2 = 0
Or, x = - 2, i.e. zero of (x + 2) = - 2
Now, (p – 2) = (- 2)^3 + 3( - 2)^2 + 3( - 2) + 1
= - 8 + 12 – 6 + 1 = - 1

Since, p(-2) ≠ 0, therefore, according to Factor Theorem g(x) = x + 2 is not the factor of given polynomial p(x) = x^3 + 3x^2 + 3x + 1

(iii) p(x) = x^3 – 4x^2 + x + 6
g(x) = x – 3

Answer: Let g(x) = 0
Or, x – 3 = 0
Or, x = 3
Now, p(3) = 3^3 – 4xx3^2 + 3 + 6
= 27 – 36 + 9 = 0

Since, p(3) = 0, therefore, according to Factor Theorem, g(x) = x + 3 is the factor of given polynomial p(x) = x^3 – 4x^2 + x + 6