Question

One of the zeros of the polynomial function is 3. f(x)=x4−x3−7x2+x+6 What is the factored form of the function? f(x)=(x−3)(x+1)(x+2)2 f(x)=(x−3)(x+1)(x−1)(x+2) f(x)=(x−3)2(x+1)(x+2) f(x)=(x−3)(x+1)2(x+2)

Answer 1

Answer:

$f(x) = (x - 3)( x - 1)(x + 1)(x + 2)$

Step-by-step explanation:

The given function is

$f(x) = {x}^{4} - {x}^{3} - 7 {x}^{2} + x + 6$

If 3 is a zero of the polynomial, then (x-3) is a factor.

We perform the long division as shown in the attachment and rewrite f(x) to get:

$f(x) = (x - 3)( {x}^{3} + 2 {x}^{2} - x - 2)$

It turns out that we can factor the resulting cubic polynomial by grouping.

$\implies \: f(x) = (x - 3)( {x}^{2}(x+ 2 ) -1 (x + 2))$

We factor further to get:

$\implies \: f(x) = (x - 3)( {x}^{2} - 1)(x + 2)$

$\implies \: f(x) = (x - 3)( x - 1)(x + 1)(x + 2)$