How do you factor completely #P(x)=x^4+4x^3-7x^2-34x-24#?

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Answer 1 · 2016-12-12T04:57:32

#P(x)=(x+1)(x+2)(x+4)(x-3)#

The number of changes in signs of the coefficients of #P(+-x)# are1

and 1 and 3 respectively. So, the number of real roots is (0+0) 0 or

(1+1) 2 or (1+3).

The sum of the coefficients in P(-x) is 0. So, -1 is a zero of P.

The graph reveals zeros near x = -4, -2,-1 and 3.

Easily, P(-2) = P(-4) = P(3) = 0.

And so,

#P(x)=(x+1)(x+2)(x+4)(x-3)#

graph{y-x^4-4x^3+7x^2+34x+24=0 [-5, 5, -2.5, 2.5]}