Question

Express the polynomial as a product of linear factors.
ƒ(x)=3x^3+12x^2+3x-18

A. (x+3)(x+6)(x-1)
B. 3(x-1)(x+3)(x+2)
C. (x-2)(x+3)(x-3)
D. (x-3)(x+3)(x-2)

Answer 1

ƒ(x)=3x^3+12x^2+3x-18 = 3(x-1)(x+3)(x+2)

Answer 2

Answer:

Option B - $f(x)=3x^3+12x^2+3x-18=3(x-1)(x+2)(x+3)$                  

Step-by-step explanation:

Given : Polynomial $f(x)=3x^3+12x^2+3x-18$

To find : Express the polynomial as a product of linear factors?

Solution :

Factor of the polynomial $f(x)=3x^3+12x^2+3x-18$

Taking 3 common

$f(x)=3(x^3+4x^2+x-6)$

Now, We factor the cubic term by rational root theorem.

If a polynomial function has integer coefficients, then every rational zero will have the form  $\frac{p}{q}$ where  p  is a factor of the constant and  q  is a factor of the leading coefficient.

$p=\pm(1,2,3,6)\\q=\pm1$

The possible roots of the polynomial function is

$\pm1,\pm2,\pm3,\pm6$

Now, we substitute the values in the polynomial if it is equal to zero then it is the root.

Substitute x=1

$f(1)=1^3+4(1)^2+1-6=0$

So, x=1 is one of the root.

Similarly we substitute all the values one by one.

The values satisfied is x=-2 and x=-3

Substitute x=-2

$f(-2)=(-2)^3+4(-2)^2-2-6=0$

So, x=-2 is one of the root.

Substitute x=-3

$f(-3)=(-3)^3+4(-3)^2-3-6=0$

So, x=-3 is one of the root.

So, The factors of $f(x)=x^3+4x^2+x-6$ is (x-1)(x+2)(x+3).

Therefore, The linear factor of the given polynomial is

$f(x)=3x^3+12x^2+3x-18=3(x-1)(x+2)(x+3)$

So, Option B is correct.