Question

$\sqrt{3 } \: and \: - \sqrt{3}$
are zeroes of
$f(x) = {x}^{3} + 13 {x}^{2} + 32x + 2$
find its other zeroes
answer fast
plz
plz
help​

Answer 1

Answer:

The all zeroes of the polynomial are -10, -2 and -1.

Step-by-step explanation:

The given polynomial is

f(x)=x^3+13x^2+32x+20f(x)=x3+13x2+32x+20

It is given that -2 is a zero of the function. It means (x+2) is a factor of given polynomial.

Divide f(x) by (x+2), to find the remaining factor.

Using long division method, we get

\frac{x^3+13x^2+32x+20}{x+2}=x^2+11x+10x+2x3+13x2+32x+20=x2+11x+10

The function can be written as

f(x)=(x+2)(x^2+11x+10)f(x)=(x+2)(x2+11x+10)

f(x)=(x+2)(x^2+10x+x+10)f(x)=(x+2)(x2+10x+x+10)

f(x)=(x+2)(x(x+10)+1(x+10))f(x)=(x+2)(x(x+10)+1(x+10))

f(x)=(x+2)(x+10)(x+1)f(x)=(x+2)(x+10)(x+1)

Equate f(x)=0, to find the zeros.

(x+2)(x+10)(x+1)=0(x+2)(x+10)(x+1)=0