1. ## Factoring...3

What is the first step in factoring the following problem?

(5a^2 - 11a + 10)^2 - (4a^2 -15a + 6)^2

2. ## Re: Factoring...3

What is the first step in factoring the following problem?

(5a^2 - 11a + 10)^2 - (4a^2 -15a + 6)^2
It's the difference of two squares.

Recall that $x^2 - y^2 = (x - y)(x + y)$.

$(5a^2 - 11a + 10)^2 - (4a^2 -15a + 6)^2 =$

$[(5a^2 - 11a + 10) - (4a^2 - 15a + 6)][(5a^2 - 11a + 10) + (4a^2 - 15a + 6)] =$

3. ## Re: Factoring...3

Originally Posted by greg1313
It's the difference of two squares.

Recall that $x^2 - y^2 = (x - y)(x + y)$.

$(5a^2 - 11a + 10)^2 - (4a^2 -15a + 6)^2 =$

$[(5a^2 - 11a + 10) - (4a^2 - 15a + 6)][(5a^2 - 11a + 10) + (4a^2 - 15a + 6)] =$
I will post my work and solution this weekend.

4. ## Re: Factoring...3

(a^2 - 26a + 16)(a^2 - 26a + 16)

a^4 - 26a^3 + 16a^2 - 26a^3 + 676a^2 - 416a + 16a^2 -

416a + 256 = a^4 - 52a^3 + 740a^2 - 832a + 256.

Is this correct?

5. ## Re: Factoring...3

Originally Posted by greg1313

$(5a^2 - 11a + 10)^2 - (4a^2 -15a + 6)^2 =$

$[(5a^2 - 11a + 10) - (4a^2 - 15a + 6)][(5a^2 - 11a + 10) + (4a^2 - 15a + 6)] =$
(a^2 - 26a + 16)(a^2 - 26a + 16)

a^4 - 26a^3 + 16a^2 - 26a^3 + 676a^2 - 416a + 16a^2 -

416a + 256 = a^4 - 52a^3 + 740a^2 - 832a + 256.

Is this correct?
No.

$(5a^2 - 11a + 10) - (4a^2 - 15a + 6) =$

$5a^2 - 11a + 10 - 4a^2 + 15a - 6 =$

$5a^2 - 4a^2 - 11a + 15a + 10 - 6 =$

$a^2 + 4a + 4$

and

$(5a^2 - 11a + 10) + (4a^2 - 15a + 6) =$

$5a^2 - 11a + 10 + 4a^2 - 15a + 6 =$

$5a^2 + 4a^2 - 11a - 15a + 10 + 6 =$

$9a^2 - 26a + 16$

Substitute them back in:

$(a^2 + 4a + 4)(9a^2 - 26a + 16) =$

Do not multiply them back together. Try to factor each trinomial further into a pair of binomial factors.

6. ## Re: Factoring...3

Originally Posted by greg1313
No.

$(5a^2 - 11a + 10) - (4a^2 - 15a + 6) =$

$5a^2 - 11a + 10 - 4a^2 + 15a - 6 =$

$5a^2 - 4a^2 - 11a + 15a + 10 - 6 =$

$a^2 + 4a + 4$

and

$(5a^2 - 11a + 10) + (4a^2 - 15a + 6) =$

$5a^2 - 11a + 10 + 4a^2 - 15a + 6 =$

$5a^2 + 4a^2 - 11a - 15a + 10 + 6 =$

$9a^2 - 26a + 16$

Substitute them back in:

$(a^2 + 4a + 4)(9a^2 - 26a + 16) =$

Do not multiply them back together. Try to factor each trinomial further into a pair of binomial factors.
I get it now. I can take it from here. I made simple errors in my quick reply. I love this stuff. Thank God for math!

7. ## Re: Factoring...3

Originally Posted by greg1313
No.

$(5a^2 - 11a + 10) - (4a^2 - 15a + 6) =$

$5a^2 - 11a + 10 - 4a^2 + 15a - 6 =$

$5a^2 - 4a^2 - 11a + 15a + 10 - 6 =$

$a^2 + 4a + 4$

and

$(5a^2 - 11a + 10) + (4a^2 - 15a + 6) =$

$5a^2 - 11a + 10 + 4a^2 - 15a + 6 =$

$5a^2 + 4a^2 - 11a - 15a + 10 + 6 =$

$9a^2 - 26a + 16$

Substitute them back in:

$(a^2 + 4a + 4)(9a^2 - 26a + 16) =$

Do not multiply them back together. Try to factor each trinomial further into a pair of binomial factors.
a^2 + 4a + 4 factors out to be (a + 2)(a + 2).

9a^2 - 26a + 16 factors out to be (3a - 2)(3a - 8).

(a + 2)(a + 2)(3a - 2)(3a - 8).

8. ## Re: Factoring...3

a^2 + 4a + 4 factors out to be (a + 2)(a + 2).

9a^2 - 26a + 16 factors out to be (3a - 2)(3a - 8). No, that should be (a - 2)(9a - 8).