Question

Two zeroes of polynomial x3-6x2+11x-6 are 2 and 3, find the third zero

Answer 1

Hi there!

Given :-
2 and 3 are the zeroes of poly. x³ - 6x² + 11x - 6

Thus,
(x - 2)(x - 3) will be a factor of p(x) = x³ - 6x² + 11x - 6

∵ Given poly. is cubic so,

x³ - 6x² + 11x - 6 = (x -2)(x - 3)(ax + b)

x³ - 6x² + 11x - 6 = (x² - 5x - 6)(ax + b)

x³ - 6x² + 11x - 6 = ax³ - 5ax² + 6ax + bx² - 5bx + 6b

x³ - 6x² + 11x - 6 = ax³ - x²(5a - b) + x(6a - 5b) + 6b

Comparing coefficients of x on both sides,

a = 1 and b = - 1

Thus, the factor (ax + b) = (x - 1) Or x = 1

Hence, The required answer is :-
The third zero is : x = 1

Hope it helps! :)

Answer 2

Answer:

The third zero is 1

Step-by-step explanation:

Given 2 and 3 are the zeroes of polynomial $x^3-6x^2+11x-6$

Thus, (x - 2)(x - 3) will be a factor of $x^3-6x^2+11x-6$

Let the third factor be (ax+b)

∵ Given polynomial is cubic so,

$x^3-6x^2+11x-6=(x -2)(x - 3)(ax + b)$

$x^3-6x^2+11x-6=(x^2-3x-2x+6)(ax + b)$

$x^3-6x^2+11x-6=(x^2-5x+6)(ax + b)$

$x^3-6x^2+11x-6=(ax^3+bx^2-5ax^2-5bx+6ax+6b)$

$x^3-6x^2+11x-6=(ax^3+(b-5a)x^2+(6a-5b)x+6b)$

Comparing coefficients on both sides

a = 1 and b = - 1

Thus, the factor (ax + b) = (x - 1)

Hence, the third zero is x = 1