How do you factor x^2 + 7x + 6
and 49x^2 - 16?
Pleasee help :)

Question

How do you factor x^2 + 7x + 6 
and 49x^2 - 16? 
Pleasee help :)

anonymous · Answer

Remember to start out with listing what the constant can be divided by: for a) 6, 1, 3, 2 and for b) 4, 4, 1, 16

anonymous · Answer

Now for a), look at the middle factor. Are there any two numbers that multiply to six and add to seven?

anonymous · Answer

Once you find that, just put it in (x+a)(x+b) form and you're done.

anonymous · Answer

The second one's a bit more complicated. There is no middle term. This is of course because it is 0. So, you need to find two numbers that multiply to -16 and add to 0.

anonymous · Answer

As well, with b), you must take in account the coefficient of x^2, which is 49.

anonymous · Answer

With this type of question, the rule of thumb is that you use the square root of the coeffienct or constant for your factoring; in this case, 7 and 4.

anonymous · Answer

So the answer would be (7x+4)(7x-4)

anonymous · Answer

(for b)

whpalmer4 · Answer

You should train yourself to look at any expression of the form - and ask yourself if they could both be squares. The difference of squares is readily factored: $$a^2- b^2 = (a+b)(a-b)$$ Here we have $49x^2-16$ so $a^2=49x^a$ and $b^2=16$ That gives us $a = 7x$ and $b = 4$, so the factoring is $$49x^2-16 = (a+b)(a-b) = (7x+4)(7x-4)$$