Question

Find the value of c so that (x+3) is a factor of the polynomial p(x)

Answer 1

Answer:

Step-by-swe know that x + 3 is a factor of the polynomial, that means that if we divide the polynomial by (x + 3), we'll get a remainder of 0.

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Explanation:

Likewise by the remainder theorem, since that x + 3 = 0, or x = -3, if we plug that -3 as the argument for p(x), namely p(-3), our output will be zero, if indeed (x + 3) is a factor of that polynomial, so then, let's plug that -3 in p(x).

Answer 2

we know that x + 3 is a factor of the polynomial, that means that if we divide the polynomial by (x + 3), we'll get a remainder of 0.

likewise by the remainder theorem, since that x + 3 = 0, or x = -3, if we plug that -3 as the argument for p(x), namely p(-3), our output will be zero, if indeed (x + 3) is a factor of that polynomial, so then, let's plug that -3 in p(x).

$\bf p(x) = x^3-4x^2+cx+33 \\\\\\ p(-3)=(-3)^3-4(-3)^2+c(-3)+33 \\\\\\ \stackrel{p(-3)}{0}=(-27)-4(9)-3c+33 \implies 0=-27-36-3c+33 \\\\\\ 0=-30-3c\implies 30=-3c\implies \cfrac{30}{-3}=c\implies -10=c \\\\[-0.35em] ~\dotfill\\\\ ~\hfill p(x) = x^3-4x^2-10x+33~\hfill$