I"ve searched almost everywhere the internet and cannot seem to factorise this polynomial.

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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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