Factorising fourth power polynomial with 5 terms

Question

I've searched all over the internet and cannot seem to factorise this polynomial.

$x^4 - 2x^3 + 8x^2 - 14x + 7$

The result should be $(x − 1)(x^3 − x^2 + 7x − 7)$

What are the steps to get to that result? I've tried grouping but doesn't seem to work...

Synthetic division allows you to quickly check for any first degree factors. — hardmath, Aug 13 '18 at 18:25
You could add up the coefficients and notice that the sum is $0$, which means that $x=1$ is a root. — Zubin Mukerjee, Aug 13 '18 at 18:25
wikipedia's explanation is very complicated.. No idea what synthetic division is. Zubin Mukerjee thanks! — Daisy123, Aug 13 '18 at 18:29
If you write the polynomial in the form $x^4 - 2x^2 + x^2 + 7x^2 - 14x + 7$, you can factor it by grouping the first three terms and last three terms. — N. F. Taussig, Aug 13 '18 at 18:31
Ive tried doing that but I get x^2 (x^2 -2x + 8) + 7 (-2x+1) and then trying to factorise (x^2 - 2x +8) doesnt work out.. as (x+4)(x-2) gives a negative 8 as if it was to give a positive 8 then you cant make -2x — Daisy123, Aug 13 '18 at 18:34

score 2 · Accepted Answer · answered Aug 13 '18 at 18:42

$x-a$ is a factor of $x^4 - 2x^3 + 8x^2 - 14x + 7$ if and only if $a^4 - 2a^3 + 8a^2 - 14a + 7=0.$ Integers that can work are the divisors of $7$ (the independent term). That is: $\pm 1,\pm 7.$ If we check with $x=1$ we get

$$1^4 - 2\cdot 1^3 + 8\cdot 1^2 - 14\cdot1 + 7=0.$$ So $x-1$ is a factor. That is, there exists a polynomial $p(x)$ of degree $3$ such that $$x^4 - 2x^3 + 8x^2 - 14x + 7=(x-1)p(x).$$

To get $p(x)$ you have to divide $(x^4 - 2x^3 + 8x^2 - 14x + 7):(x-1)$ in the way you prefer.

You should get

$$(x^4 - 2x^3 + 8x^2 - 14x + 7)=(x-1)(x^3-x^2+7x-7).$$

Proceeding in the same way you have

$$x^3-x^2+7x-7=(x-1)(x^2+7).$$ Or

$$(x^4 - 2x^3 + 8x^2 - 14x + 7)=(x-1)^2(x^2+7).$$

yakobyd · Answer 2 · 2018-08-13T18:55:41.283

1

Considering $(a-b)^2 = a^2 - 2ab + b^2$, notice that you may manipulate the polynomial as follows:

\begin{align} x^4 - 2x^3 + 8x^2 -14x + 7 &= (x^4 -2x^3 + x^2) + (7x^2 - 14x + 7) \\ &= x^2(x^2 - 2x + 1) + 7(x^2 - 2x + 1) \\ &= (x-1)^2(x^2 + 7) \end{align}

which yields a complete factorization over the real numbers.

edited Aug 13 '18 at 18:55

answered Aug 13 '18 at 18:34

yakobyd

681

I need this result (x−1)(x3−x2+7x−7) – Daisy123 Aug 13 '18 at 18:37
You have that $$(x-1)^2(x^2 + 7) = (x-1)(x-1)(x^2 + 7) = (x-1)(x^3 - x^2 + 7x - 7)$$ – yakobyd Aug 13 '18 at 18:38
x2(x2−2x+1)+7(x2−2x+1) no idea how you get this – Daisy123 Aug 13 '18 at 18:45
I have added an additional step. I hope it is clearer now. – yakobyd Aug 13 '18 at 18:56

score 1 · Answer 3 · answered Aug 13 '18 at 18:40

1

As $p(1)=0$, you know that $x-1$ is a factor. Now

$$x^4 - 2x^3 + 8x^2 - 14x + 7 \\=x^3(x-1)-x^3+8x^2-14x+7 \\=x^3(x-1)-x^2(x-1)+7x^2-14x+7 \\=x^3(x-1)-x^2(x-1)+7x(x-1)-7x+7 \\=x^3(x-1)-x^2(x-1)+7x(x-1)-7(x-1).$$

answered Aug 13 '18 at 18:40

x3(x−1)−x3+8x2−14x+7 where did the 2 dissapear? – Daisy123 Aug 13 '18 at 18:44
@NadiaKuznecova Note that $x^3(x-1)-x^3=x^4-x^3-x^3=x^4-2x^3.$ – mfl Aug 13 '18 at 18:49

Factorising fourth power polynomial with 5 terms

3 Answers3