Question

Given that one zero is –4, find all zeros of P(x) = x3 + x2 – 22x – 40.

AThe roots of the equation are –5, –4, and –2.
B The roots of the equation are 7, –4, and 3.
C The roots of the equation are 10, –4, and 2.
D The roots of the equation are 5, –4, and –2.

Answer 1

You can divide the right side of P(x) by (x + 4) to get a quadratic.

Answer 2

that should be x^3 and x^2

Answer 3

Yes, I saw that and accounted for that in my answer above. So, just divide by that factor above.

Answer 4

You can divide by either synthetic division or long division. You'll get the same quadratic.

Answer 5

Are you familiar with synthetic division?

Answer 6

and now the remaining factor is x^2 -3x -10 which can easily be factored.

Answer 7

So you'll have factors (x + 4)(x - 5)(x + 2)

Answer 8

Gives zeros of -4, 5, and -2.

Answer 9

lets do synthetic since its (or at least i find it) easier :)

\[\underline{-4~|}~~~~1~~~~~~1~~~-22~~~-40\]\[~~~~~~~~\underline{~~~~~~-4~~~~~~~12~~~~~~~~40}\]\[~~~~~~~~~~1~~-3~~~-10~~~~~~~~~0\]so now we have: \(x^2-3x-10\) wich will be factored:\[(x-5)(x+2)\]set those equal to zero and find the zeros:\[x-5=0~~~\implies~~~x=5\]\[x+2=0~~~\implies~~~x=-2\]
so the three zeros of \(P(x) = x^3 + x^2 – 22x – 40\) are \(-4, ~-2,~and~5\)

Answer 10

hope that's helpful @stevench ! :)