Factorise : x³ - 6x² +11x - 6

Solution:

Given, the polynomial is x³ - 6x² +11x - 6

Let p(x) = x³ - 6x² +11x - 6

The constant term of p(x) is -6.

Factors of -6 = ±1, ±2, ±3, ±6

Let us take x = 1

Substitute x = 1 in p(x),

p(1) = (1)³ - 6(1)² +11(1) - 6

= 1 - 6 + 11 - 6

= 1 + 11 - 6 - 6

= 12 - 12

= 0

So, x - 1 is a factor of x³ - 6x² + 11x - 6.

Now splitting the x² and x terms,

x³ - 6x² - 11x - 6 = x³ - x² - 5x² + 5x + 6x - 6

Taking (x - 1) as a common factor,

= x²(x - 1) - 5x(x - 1) + 6(x - 1)

= (x² - 5x + 6)(x - 1)

On factoring x² - 5x + 6 by splitting the middle term,

= x² - 3x - 2x + 6

= x(x - 3) - 2(x - 3)

= (x - 2)(x - 3)

Therefore, the factors of p(x) are (x - 2)(x - 3)(x - 1)

✦ Try This: Factorise : x³ - 3x² + 17x - 8

☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 2

NCERT Exemplar Class 9 Maths Exercise 2.3 Problem 24(ii)

Summary:

Trinomial is a type of polynomial that has three terms. The factors of x³ - 6x² - 11x - 6 are (x - 2)(x - 3)(x - 1)

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