2021-12-19

All the real zeros of the given polynomial are integers. Find the zeros, and write the polynomial in factored form.
$P\left(x\right)={x}^{3}-4{x}^{2}-19x-14$

Durst37

Expert

Step 1
Given polynomial is, $P\left(x\right)={x}^{3}-4{x}^{2}-19x-14$
In factored form, it is written as follows:
${x}^{3}-4{x}^{2}-19x-14={x}^{3}+{x}^{2}-5{x}^{2}-5x-14x-14$
$={x}^{2}\left(x+1\right)-5x\left(x+1\right)-14\left(x+1\right)$
$=\left(x+1\right)\left({x}^{2}-5x-14\right)$
=(x+1)(x(x+2)-7(x+2))
=(x+1)(x+2)(x-7)
Step 2
It zeroes are find by putting it equal to zeros.
${x}^{3}-4{x}^{2}-19x-14=0$
(x+1)(x+2)(x-7)=0
x+1=0 or x+2=0 or x-7=0
x=-1, x=-2, x=7

Ella Williams

Expert

$P\left(x\right)={x}^{3}-4{x}^{2}-19x-14$
rational roots of the function are
x=+-{1,2,7,14}
actual roots occur at
x = -1 , -2 , 7
hence zeros are
x = -1 , -2 , 7
polynomial in factored form is
P(x) = (x+1) ( x+2) (x-7)

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