Question

Find all the zeroes of the polynomial $p\left(x\right)=x^4+x^3-34x^2-4x+120$ if two of its zeroes are $2$ and $-2.$

Answer 1

Step 1: Find the factor using zeros of the polynomial and find the divisor.

2 and -2 are the zeros of the given polynomial.
Hence, (x – 2) and (x – (-2)) will be the factors of the polynomial

g(x) = (x – 2)(x + 2)
g(x) = x^2 – 4

Step 2: Apply the division Algorithm. Divide the highest degree term of the dividend by the highest degree term of the divisor.

here $x^4$ is the highest term in dividend and $x^2$ is the highest term in divisor and we will divide $x^4$ by $x^2$, we get the first term of the quotient as $x^2$

Step 3: Multiply the quotient with the divisor.

multiply the quotient obtained in the previous step with divisor, hence the product is $x^4-4x^2$

Step 4: Subtract the product of the divisor and the quotient from the dividend.

we get $x^3-30x^2-4x+120$
after subtraction

Repeat the steps 2 to 4 till the remainder is zero or deg r(x) < deg g(x).

So, the quotient is $x^2+x-30$ and the remainder is 0.

Step 5: Equate the quotient with 0 and factorize the quadratic polynomial to find the remaining zeroes.

$x^2+x-30=0$
$x^2-5x+6x-30=0$
$x(x-5)+6(x-5)=0$
$(x+6)(x-5)=0$
$x=-6,5$

So the zeros of the polynomial are 2, −2, 5 and −6.