Factor these three?

Question

Out of 13 problems, I can't solve these three.

x^4 + 15x^2 – y^4 + 13y^2 + 99
30x^2 – 7x + 120b^2 – 14b + 120xb – 1
8x^3 – 12x^2 + 2x + 7

Thanks to pete for the 2nd problem.:) As for the 1st and last prob, I found a way to make it look factorable. But I don't know how to finish the factoring.

So here goes.

x^4 + 15x^2 - y^4 + 13y^2 + 99
(x^4 + 15x^2 + (15/2)^2) - (y^4 + 13y^2 + (13/2)^2) + 1/2
(x^2+(15/2))^2 - (y^2+(13/2))^2 + 1/2
then?

8x^3 - 12x^2 + 2x + 7 + 1 - 1
(8x^3 - 12x^2 + 2x + 8) - 1

then the last question should be (8x^3 - 12x^2 + 2x + 8) instead.

Answer 1

x^4 + 15x^2 – y^4 + 13y^2 + 99
I don't see a way to factor this one either...

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30x^2 – 7x + 120b^2 – 14b + 120xb – 1
= (30x^2 + 120xb + 120b^2) - 7x - 14b - 1
= 30(x^2 + 4xb + 4b^2) - (7x + 14b) - 1
= 30(x + 2b)(x + 2b) - 7(x + 2b) - 1
= (x + 2b)[(30(x + 2b) - 7] - 1

It's not pretty, but it's the only way I can see it factoring...
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8x^3 – 12x^2 + 2x + 7
= 4x^2(2x - 3) + 2x + 7

That's about all you can do to this one...

Answer 2

Let's tackle your last one:
We must check for rational zeros and use the fact that
if r is a rational zero, then x-r is a factor.
Also, any rational zero must divide 7.
So the only possibilities are-1,1,-7 and 7.
Trying all these, we see that none is a zero of
8x^3 – 12x^2 + 2x + 7, so it's irreducible.

Answer 3

I'll just finish pete's answer because his answer are not factors yet.

His last statement was:
(x + 2b)[(30(x + 2b) - 7] - 1

You bring back the (x+2b) inside.

Let a=x+2b

30a^2- 7a - 1
(3a - 1)(10a + 1)

then substitute the x+2b to the a then you'll get the factors.