Question

use factor theorem to factorise each of the following polynomials x³-x²-14x+24​

Answer 1

Answer:

x3+x2−14x−24

Use the monomial −14x

It is equal to −4x−10x

x3+x2−4x−10x−24

Rearrange

x3−4x+x2−10x−24

Regroup

(x3−4x)+(x2−10x−24)

Factoring

x(x2−4)+(x+2)(x−12)

x(x+2)(x−2)+(x+2)(x−12)

Factor out the common binomial factor (x+2)

(x+2)[x(x−2)+(x−12)]

Simplify the expression inside the grouping symbol [ ]

(x+2)[x2−2x+x−12]

(x+2)(x2−x−12)

Factoring the trinomial x2−x−12=(x+3)(x−4)

We now have the factors

(x+2)(x+3)(x−4)

Final answer

x3+x2−14x−24=(x+2)(x+3)(x−4)

God bless ....I hope the explanation is useful.

Answer 2

Step-by-step explanation:

The factor theorem states that if x−a is a factor of some polynomial f(x), then f(a)=0. The proof of this is very straightforward. Try it as an exercise.

To find factors of your cubic,

f(x)=x3−x2−14x+24

We will try to find values of x for which f(x)=0. Supposing that the polynomial can be neatly factored into integer factors x−a, x−b, and x−c, we know that abc=24, so it is beneficial to try positive and negative factors of 24.

Let's try 1, the smallest:

f(1)=13−12−14⋅1+24=10

Hmm, no luck. Next let's try −1:

f(1)=(−1)3−(−1)2−14(−1)+24=36

Again we have not reached zero. Next let's try 2:

f(2)=23−22−14⋅2+24=0

We've found a solution! This means that x−2 is a factor of the cubic. To find the remaining factors, we can do polynomial division to find some quadratic x2+bx+c so that f(x)=(x−2)(x2+bx+c).

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