Question

Factorise: x³ - 13x - 12​

Answer 1

Answer:

VERIFIED ANSWER

Step-by-step explanation:

Given

f(x)=x3−13x−12

To solve this type of problem, We have to find out at least one root by hit and trial using the factor theorem.

Here is a trick:

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

Here

p=±1,±2,±3,±4,±6,±12 and 

q=±1

Find every combination of ±qp. 

These are the possible roots of the polynomial function.

±1,±2,±3,±4,±6,±12

Substituting x=1 in f(x), we get

f(−1)=(−1)3−13(−1)12=−1+13−12=13+13=0

∴(x+1) is a factor of f(x)  

Now, dividing f(x) by (x+1), we get

x3−13x−12=(x+1)(x2−x−12)

=(x+1)(x2−4x+3x−12)

=(x+1){x(x−4))+3(x−4)}

=(x+1)(x+3)(x−4)

Answer 2

Answer:

Given

f(x)=x³ −13x−12

To solve this type of problem, We have to find out at least one root by hit and trial using the factor theorem.

Here is a trick: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.

Here

p=±1,±2,±3,±4,±6,±12 and

q=±1

Find every combination of ±

$\frac{p}{q}$

.

These are the possible roots of the polynomial function.

±1,±2,±3,±4,±6,±12

Substituting x=1 in f(x), we get

f(−1)

=(−1) ³ −13(−1)12

=−1+13−12

=13+13

=0

∴(x+1) is a factor of f(x)

Now, dividing f(x) by (x+1), we get

x³−13x−12=(x+1)(x²−x−12)

=(x+1)(x

−4x+3x−12)

=(x+1){x(x−4))+3(x−4)}

=(x+1)(x+3)(x−4)