Given P(x) = x^3 - 3x^2 + 16x - 48, find the EXACT Zeros of P by factoring:

Question

Given P(x) = x^3 - 3x^2 + 16x - 48, find the EXACT Zeros of P by factoring:

The Zeros are: 3, 4i, -4i.

Write P in factored form (as a product of linear factors). Be sure to write the full equation, including P(x)=.

P(x)=_______________________

Answer 1

Here we can factor by grouping. Lets group the first two terms together and the last 2 terms together, then distribute out a GCF out of each group.

x³ - 3x^{2 +} 16x - 48

(x³ - 3x²) ⁺ (16x - 48)

x² (x-3) + 16(x-3)

Afterwards, we can take out the x-3 and factor the difference of squares

(x-3)(x²+16)

(X-3)

x= 3

For the quadratic, set it equal to zero to solve and when you subtract both sides by 16 and take the square root, you will realize you cannot square root a negative number to get a real answer, therefore the 4i and -4i are imaginary numbers.

plug the numbers to the original equation double check your answer.

Answer 2