Question

Factor expression below. Write as a polynomial in descending order.
16x^2-1

Answer 1

That's a difference of two squares:

\[(a-b)(a+b) = a^2-ba+ba-b^2 = a^2-b^2\]

Answer 2

still confused....

Answer 3

Well, what is the square root of 1?

Answer 4

1

Answer 5

Good. What is the square root of \(16x^2\)?

Answer 6

4x??

Answer 7

Yes! So now do you see why I called it a difference of two squares?

Answer 8

It's \((4x)^2 - (1)^2\), right?

Answer 9

could u help me with 9 more questions?? I need them to graduate in 24 hours...

Answer 10

If you have a difference of two squares, \(a^2-b^2\), you can factor that as
\[(a+b)(a-b)\]
Here we have \(a=4x\) and \(b = 1\) so it factors as
\[(16x^2-1) = (4x-1)(4x+1)\]

Answer 11

What's the next question?

Answer 12

left side of the attachment

Answer 13

Here we've got a difference of squares going in the other direction: we've got the factors, and we'll get the difference of squares.

\[\frac{x+3}{x+2}*\frac{x-3}{x-2}\]What does the numerator become after multiplying? How about the denominator? Remember, \((a+b)(a-b) = a^2-b^2\)

Answer 14

sooo B??

Answer 15

yes!

Answer 16

1 more, then I have to go...

Answer 17

okay sounds good left side again

Answer 18

A couple of ways you could go here. Just read intersection point off the graph, or set the two equations equal to each other and solve for \(x\). \[\frac{3}{5}x-1 = x-3\]Multiply through by 5\[3x-5=5x-15\]Subtract 3x from both sides, add 15 to both sides:\[10=2x\]\[x=5\]now plug that into either formula to get \(y\)\[y=x-3=5-3=2\]so our point of intersection is \((5,2)\) and if you look at the graph, I think you'll agree that is where the lines intersect.

Answer 19

thank you so much u have been a great help!

Answer 20

Hopefully if you've got any more of those types of problems you won't have any trouble following the example...good luck!