Question

Factorise
$6x {}^{3} - 25x {}^{2} + 32x - 12$

Answer 1

Final Answer : (3x-2)(2x-3)(x-2)

Let f(x) =
$6 {x}^{3} - 25 {x}^{2} + 32x - 12$

Steps:
1) By inspection, we observe that
f(2) = 6(2)^3 -25(2)^2 +32(2) -12
= 48-100+64-12 = 0
=> (x-2) is factor of given polynomial.

2) Now, we divide the given polynomial by (x-2).
We get quotient as :
$6 {x}^{2} - 13x + 6$
For Calculation see pic.

3)
$6 {x}^{2} - 13x + 6\\ = > 6 {x}^{2} - 9x - 4x + 6 \\ = > 3x(2x - 3) - 2(2x - 3)\\ = > (2x - 3)(3x - 2)$
4) Now,
$6 {x}^{3} - 25 {x}^{2} + 32x - 12 = \\ (x - 2)(2x - 3)(3x - 2)$

Answer 2

$\textbf{\huge\underline{\underline\red{Solution} : - }} \\ \\ \bold{ \underline\red{let }} \\ \\\bold{\purple {f(x ) =6{x}^{3} - 25 {x}^{2} + 32x - 12 }} \\ \\ \bold{ \underline\red {now \: take \: x \: = \: 2}} \\ \bold{ \underline\red{so}} \\ \\\bold{\purple {f(x = 2) =6{(2)}^{3} - 25 {(2)}^{2} + 32(2) - 12 }} \\ \\\bold{\purple {f(x = 2) =6(8) - 25 (4) + 64 - 12 }} \\ \\\bold{\purple {f(x = 2) =48 - 100 + 52 }} \\ \\\bold{\purple {f(x = 2) = - 52 + 52 }} \\ \\ \bold{\purple {f(x = 2) = 0 }} \\ \\ \bold{ \underline\blue{on \: puting \: x \: = \: 2 \: \:we \: get \: f(x) = \: 0} }\\ \bold{ \underline\blue{hence \: one \: of \: the \: factor \: of \: given \: }} \\ \bold{ \underline\blue{polynomial \: is \: (x - 2).}} \\ \\ \bold{\underline \red{now}}\\ \\\bold{\purple { =>6{x}^{3} - 25 {x}^{2} + 32x - 12 }} \\ \\ \bold{\purple { =( x - 2)(6 {x}^{2} - 13x + 6 )}} \\ \\ \bold{\purple {=( x - 2)(6 {x}^{2} - 9x - 4x + 6 )}} \\ \\ \bold{\purple { =( x - 2)(3x(2x - 3) - 2( 2x - 3))}} \\ \\ \bold{\purple { =( x - 2)(2x- 3)( 3x - 2)}} \\ \\ \fbox{\bold{ \underline\red{i \: hope \: it \: helps \: you.}}}$