Question

Factorize using the factor theorem 2x^3− 13x^2+6x+45

Answer 1

Answer:

let f(x)=2x^3-13x^2+6x+45

for x=3,

f(3)=2(3)^3-13(3)^2+6(3)+45

=54-117+18+45

=117-117

=0

hence x-3=0 is the factor of f(x)

Now,

a/c to factor th.

x-3)2x^3-13x^2+6x+45(2x^2-7x-15

2x^3-6x^2

-. +

_______________

0. -7x^2+6x

-7x^2+21x

+. -

______________

0. -15x+45

-15x+45

+. -

_______________

0

now f(x)=(x-3)(2 x^2-7x-15)

=(x-3)(2x^2-10x+3x-15)

=(x-3){2x(x-5)+3(x-5)}

=(x-3)(x-5)(2x+3)

Answer 2

$\huge{\boxed{\red{\bf{Solution:-}}}}$

Let ${f(x)} = {2x}^{3} - {13x}^{2} + {6x} + {45}$

then,

let ${x} = {3}$

${f(3)} = {2×3}^{3} - {13×3}^{2} + {6×3} + {45}$

${f(3)} = {54} - {117} + {18} + {45}$

${f(3)} = {117} - {117}$

${f(3)} = {0}$

(x-3) is a factor of f(x).

Dividing f(x) from (x+1), we get:-

${2x}^{2} - {7x} - {15}={0}$

Factorizing ${2x}^{2} - {7x} - {15}={0}$:-

${2x}^{2} - {10x} + {3x} - {15}={0}$

${2x(x-5)-3(x-5)}={0}$

${(2x-3)(x-5)}={0}$

$\huge{\boxed{\red{\bf{Answer:-}}}}$

Hence the factors of f(x) are:-

(x - 5)(2x - 3)(x - 3)