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Mathematics 19 Online
OpenStudy (anonymous):

What is the factored form of b(2) + 12b + 32?

OpenStudy (luigi0210):

Do you mean: \(b^2+12b+32\)?

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

(b-8) (b-4)

OpenStudy (anonymous):

This is a FOIL problem in reverse.\[ b^2 + 12b + 32=(b+8)(b+4)\]

OpenStudy (luigi0210):

You guys might want to expand on that.

OpenStudy (anonymous):

The eight and four multiply to 32 and add to 12.

OpenStudy (anonymous):

how do I solve this not just the answer

OpenStudy (anonymous):

ok thankyou

OpenStudy (luigi0210):

@AnimalAin Just gave a small explanation on how.

OpenStudy (anonymous):

can you guys help me on another one

OpenStudy (anonymous):

Sure.

OpenStudy (anonymous):

What is the factored form of 4k2 + 20k + 25?

OpenStudy (anonymous):

This one is a little more complicated. We need to use the ac method, since the leading coefficient (in this case, four) isn't one. Remember our general representation of a quadratic function y = ax^2 + bx + c First, find the product of a and c. In this problem, that would be four times twenty-five, which equals 100. Now find the factorization of 100 that adds to twenty. I think it is ten and ten ( 10 times 10 is 100; 10 + 10 = 20). Rewrite the expression as follows:\[4k^2 + 20k + 25=4k^2 + 10k+10k + 25\]Now factor by grouping.

OpenStudy (anonymous):

\[4k^2 + 20k + 25=4k2 + 10k+10k + 25=2k(2k +5)+5(2k +5)=(2k+5)^2\]

OpenStudy (anonymous):

Actually, if I had been more observant, I would have recognized that it was a perfect square at the start, and saved myself a bunch of work. Live and learn...

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