Question

using remainder theorem, factorise: 6x^3-25x^2+32x-12

Answer 1

f(x) = 6x³-25x²+32x-12    on putting x = 2 we get f(x) = 0 , hence (x-2) is a factor of given polynomial.
(x-2)(6x²-13x+6)
= (x-2)(6x²-9x-4x+6)
= (x-2){3x(2x-3)-2(2x-3)}
= (x-2)(2x-3)(3x-2)

Answer 2

Factors are $\bf \red{6 {x}^{3} - 25 {x}^{2} + 32x - 12 = (x - 2)(2x - 3)(3x -2 )}$

Given:

• $6 {x}^{3} - 25 {x}^{2} + 32x - 12$

To find:

• Find the factors using remainder theorem.

Solution:

Theorem to be used:

Remainder Theorem:If (x-a) is a factor of polynomial p(x) then p(a)=0.

Step 1:

Find first factor using remainder theorem.

put x= 2

Let the polynomial is p(x).

$p(2) = 6 {(2)}^{3} - 25 {(2)}^{2} + 32(2) - 12$

or

$p(2) = 6 \times 8 - 25 \times 4 + 64 - 12$

or

$p(2) = 48 - 100 + 64 - 12$

or

$p(2) = 112 - 112$

or

$\bf p(2) = 0$

So,

$\bf (x - 2)$ is a factor of polynomial.

Step 2:

Divide p(x) by (x-2).

$\: \: \: \: \: \: \: \: \: x - 2 \: ) \: 6 {x}^{3} - 25 {x}^{2} + 32x - 12 \: (6 {x}^{2} - 13x + 6 \\ 6 {x}^{3} - 12 {x}^{2} \: \: \: \: \: \: \: \: \\( - ) \: \: \: \: \: \: ( + ) \: \: \: \: \: \: \\ - - - - - - - - - \\ - 13 {x}^{2} + 32x \\ - 13 {x}^{2} + 26x \\ ( + ) \: \: \: \: (- ) \\ - - - - - - - - \\ 6x - 12 \\ 6x - 12 \\ ( - ) \: \: \: ( + ) \\ - - - - - - \\ 0 \\ - - - - - -$

Here,

Quotient is $6 {x}^{2} - 13x + 6$

Step 3:

Factorise the quotient polynomial to find the other factors of p(x).

$6 {x}^{2} - 13x + 6 = 6 {x}^{2} - 9x - 4x+ 6$

or

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

or

$6 {x}^{2} - 13x + 6= (2x - 3)(3x - 2)$

Thus,

Factors are $\bf 6 {x}^{3} - 25 {x}^{2} + 32x - 12 = (x - 2)(2x - 3)(3x -2 )$

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