Best Answer

a^(2)- 2ab - 15b^(2)

When the first term has a coefficient of '1' (a^(2)), then look at the third term.

It is '15'. So we need two numbers that multiply to '15' and 'add/subtract'to '2'.

They are 3. & 5.

So writing up our brackets

( a 5b)(a 3b)

Which signs???

We note from the quadraic , that '15' is negative(-) . So the brackets need one positive(+) and one negative(-)

Since the middle term is also negative(-) , then the larger numerical number takes the negative.

Hence

( a - 5b)(a + 3b)

Done!!!!

Lvl 15

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a2 - 2ab - 15b2 = (a - 5b)*(a + 3b)

Q: What is a2 - 2ab - 15b2 factored completely?

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a3-4a = a(a2-4) when factored

(a2+2b2-2ab)(a2+2b2+2ab)

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(a-b)2 = a2 _ 2ab+b2

No. If you expand (a + b)2 you get a2 + 2ab + b2. This is not equal to a2 + b2

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a3-4a = a(a2-4) when factored

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A trinomial is perfect square if it can be factored into the form (a+b)2 So a2 +2ab+b2 would work.

Yes, as long as either number is 0. a2 + b2 = (a+b)2 a2 + b2 = a2 + 2ab + b2 0 = 2ab

There are 3 main rules for expansion of algebraic expressions. They are as follows: 1) a2 _ b2 = (a-b) (a+b) 2) (a+b)2 = a2 + 2ab +b2 3) (a-b)2 = a2 - 2ab +b2

I'm going to assume a few things... 1.) the 2s after a letter are 2 2.) this is addition, because it's easier for me and if you wanted another operation, you should have stated so. 2a2b+a2+5ab+3ab2+b2+2a2b+2ab 4ab+a2+5ab+3b2a+b2+4ab+2ab 15ab+3b2a+a2+b2

let binomial be (a + b)now (a+b)3 will be (a+b)(a+b)2 = (a+b)(a2 + 2ab+ b2) = a(a2+ 2ab+ b2) + b(a2 + 2ab+ b2) = a3+ 2a2b+ ab2 + a2b + 2ab2 + b3 = a3+ 2a2b+ ab2 + a2b + 2ab2 + b3 = a3 +3a2b + 3ab2 +b3 hope it helped... :D

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perfect trinomial square?? it has the form: a2 + 2ab + b2

Some special cases that are relevant in practice are: (a + b)2 = a2 + 2ab + b2 (a - b)2 = a2 - 2ab + b2 (x + a)(x + b) = x2 + (a+b)x + ab

(a-b)2 = a2 _ 2ab+b2

a2-5a-14 = (a-7)(a+2) when factored