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

The an outcome should it is in $(x − 1)(x^3 − x^2 + 7x − 7)$

What room the measures to gain to the result?I"ve make the efforts grouping however doesn"t seem come work...

$x-a$ is a variable of $x^4 - 2x^3 + 8x^2 - 14x + 7$ if and also only if $a^4 - 2a^3 + 8a^2 - 14a + 7=0.$ Integers that deserve to work space the divisors the $7$ (the independent term). That is: $pm 1,pm 7.$ If we examine with $x=1$ us get

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

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

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You should gain

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

Proceeding in the same way you have

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

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

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answered Aug 13 "18 in ~ 18:42

mflmfl

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

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

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reply Aug 13 "18 in ~ 18:40

user65203user65203

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

eginalign 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)endalign

which returns a finish factorization end the real numbers.

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edited Aug 13 "18 at 18:55

answered Aug 13 "18 in ~ 18:34

yakobydyakobyd

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