By factor theorem, show that x - 1 is a factor of x 3 - 23 x 2 + 142 x - 120 . Find the other factors of the given polynomial.

We have p x = x 3 - 23 x 2 + 142 x - 120 Factor Theorem: If p x is a polynomial of degree ≥ 1 and α is real number, then x - α is a factor of p x if p α = 0 and conversely, p α = 0 if x - α is a factor of p x . If x - 1 is a factor of p x , then by factor theorem, p 1 = 0 . Now, p 1 = 1 3 - 23 1 2 + 142 1 - 120 p 1 = 1 - 23 + 142 - 120 p 1 = 120 - 120 = 0 ⇒ x - 1 is a factor of x 3 - 23 x 2 + 142 x - 120 . To find the other factors of x 3 - 23 x 2 + 142 x - 120 , we need to consider the factors of - 120 . p 10 = 10 3 - 23 10 2 + 142 10 - 120 p 10 = 1000 - 2300 + 1420 - 120 = 0 ⇒ x - 10 is a factor of x 3 - 23 x 2 + 142 x - 120 . [By factor theorem] p 12 = 12 3 - 23 12 2 + 142 12 - 120 p 12 = 1728 - 3312 + 1704 - 120 = 0 ⇒ x - 12 is a factor of x 3 - 23 x 2 + 142 x - 120 . So, x 3 - 23 x 2 + 142 x - 120 = x - 1 x - 10 x - 12 Therefore, the other factors of x 3 - 23 x 2 + 142 x - 120 are x - 10 , x - 12 .

E M B I B E

CBSE Syllabus Standard Mathematics for Class IX>Chapter 2 - Polynomials>Exercise 2.6>Q 1

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9th CBSE

IMPORTANT

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If $(3 x - 2)$ is a factor of $3 x^{3} + x^{2} - 20 x + 12$ , find the other factors.

Important Questions on Polynomials

MEDIUM

9th CBSE

IMPORTANT

CBSE Syllabus Standard Mathematics for Class IX>Chapter 2 - Polynomials>Exercise 2.6>Q 2

By factor theorem, show that $(3 x^{3} + x^{2} - 20 x + 12) \div (3 x - 2) = x^{2} + x - 6$ is a factor of $x^{3} - 23 x^{2} + 142 x - 120$ . Find the other factors of the given polynomial.