. Denise has multiplied (2r – 3)(2r + 3) to get 4r2 – 9. After looking at her work, Hyder says her solution must be incorrect because multiplying two binomials will always produce a trinomial. Which student is correct? Explain the error to the incorrect student. HELP ME PLEASE NEED IT IN TIME THANKS.

The answer is 4r² – 9

Denise answer is correct.

Step-by-step explanation:

One of the methods of solving the given equation [(2r – 3)(2r + 3)] is the FOIL method

Using the FOIL method (Used for multiplying binomials)

F stands for first

O stands for outer

I stands for inner

L stands for last

In this method, it means we will first multiply the first terms followed by multiplying the outer term, then multiplying the inner terms and finally, the last terms.

So, for the equation  (2r – 3)(2r + 3)

First : (2r)(2r) = 4r²

Outer : (2r)(+3) = 6r

Inner : (–3)(2r) =  – 6r

Last :  (–3)(+3) =  – 9

Adding the results we get

4r² + 6r – 6r – 9 (+6r and – 6r cancels out)

= 4r² – 9

∴ (2r – 3)(2r + 3) = 4r² – 9

Hence, Denise answer is correct

x= -3 and

step-by-step explanation:

to find the roots of a cubic, use division or factoring to find the factors then set equal to 0. as this is a difference of cubes, it factors into

where a is the cube root of .

here   this is 27 and the cube root of 27 is 3. so we write using the form:

.

set each factor to 0 and solve for x.

x-3=0

x=3

the quadratic expression does not factor and must be solved using the quadratic formula.

where a = 1, b=3, and c=9

this cubic has two complex roots.