given the following polynomial ,
$x^5 −20x^3 +37x^2 +12x−44$
Is it possible to factor it using the Vieta's Formula ?
If our polynomial has integer root then $44$ is divisible by these roots.
Easy to see that $2$ and $-1$ they are roots and we can get the following factoring.
$$x^5 −20x^3 +37x^2 +12x−44=$$ $$=x^5+x^4-x^4-x^3-19x^3-19x^2+56x^2+56x-44x-44=$$ $$=(x+1)(x^4-x^3-19x^2+56x-44)=$$ $$=(x+1)(x^4-2x^3+x^3-2x^2-17x^2+34x+22x-44)=$$ $$=(x-2)(x+1)(x^3+x^2-17x+22)=$$ $$=(x-2)(x+1)(x^3-2x^2+3x^2-6x-11x+22)=$$ $$=(x-2)^2(x+1)(x^2+3x-11).$$
HINT: $x=2$ and $x=-1$ are integer solutions
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If our polynomial has integer root then $44$ is divisible by these roots.
Easy to see that $2$ and $-1$ they are roots and we can get the following factoring.
$$x^5 −20x^3 +37x^2 +12x−44=$$ $$=x^5+x^4-x^4-x^3-19x^3-19x^2+56x^2+56x-44x-44=$$ $$=(x+1)(x^4-x^3-19x^2+56x-44)=$$ $$=(x+1)(x^4-2x^3+x^3-2x^2-17x^2+34x+22x-44)=$$ $$=(x-2)(x+1)(x^3+x^2-17x+22)=$$ $$=(x-2)(x+1)(x^3-2x^2+3x^2-6x-11x+22)=$$ $$=(x-2)^2(x+1)(x^2+3x-11).$$