Question

The polynomial p(x)=x^3+7x^2-36p(x)=x 3 +7x 2 −36p, left parenthesis, x, right parenthesis, equals, x, cubed, plus, 7, x, squared, minus, 36 has a known factor of (x+3)(x+3)left parenthesis, x, plus, 3, right parenthesis. Rewrite p(x)p(x)p, left parenthesis, x, right parenthesis as a product of linear factors. p(x)=

Answer 1

x^3 + 7x^2 -36 = (x + 3)(x-2)(x + 6)

Explanation:

Here we want to rewrite the polynomial

P(x) = x^3 + 7x^2 - 36

As a product of linear factors given that we have one of the roots already as (x + 3)

What we shall do here is to divide the polynomial by x + 3

Please check attachment for long polynomial division

From the division, the other factor is x^2 + 4x -12

So we can rewrite the polynomial as;

x^3 + 7x^2 -36 = (x + 3)(x^2 + 4x -12)

So let’s now write x^2 + 4x -12 as a product of its linear factors

x^2 + 4x -12 = x^2 + 6x -2x -12

= x(x + 6)-2(x + 6)

so we have;

(x -2)(x + 6)

Therefore;

x^3 + 7x^2 -36 = (x + 3)(x-2)(x + 6)

Answer 2

(x-2)(x+3)(x+6)

Explanation:

Given the polynomial function p(x)=x^3+7x^2-36

We are to write it as a product of its linear factor

Assuming the value of x that will make the polynomial p(x) to be zero

Let x = 2

P(2) = 2³+7(2)²-36

P(2) = 8+7(4)-36

P(2) = 8+28-36

P(2) = 0

Since p(2) = 0 hence x-2 is one of the linear factors

Also assume x = -3

P(-3) = (-3)³+7(-3)²-36

P(-3) = -27+7(9)-36

P(-3) = -27+63-36

P(-3) = 36-36

P(-3) = 0

Since p(-3) = 0, hence x+3 is also a factor

The two linear pair are (x-2)(x+3)

(x-2)(x+3) = x²+3x-2x-6

(x-2)(x+3) = x²+x-6

To get the third linear function, we will divide x^3+7x^2-36 by x²+x-6 as shown in the attachment.

x^3+7x^2-36/x²+x-6 = x+6

Hence the third linear factor is x+6

x^3+7x^2-36 = (x-2)(x+3)(x+6)