Can anyone help me factor these?

Question

I don't know how to factor when there are two variables.

4m^ - 28mn +49n^

2x^3 +4x^ - 18x - 36

Answer 1

when you have two variables there should be one in front and one in the back like (m+n)(m-n). Factor like normal.

(2m-7n)(2m-7n)

the second one you split up into two pairs. group together the first two and the last two.

(2x^3 +4x^2)+(-18x-36)

Then factor out what they have in common:

2x^2(x+2)-18(x+2)

You see that we have two (x+2) which is good. put the others together:

(2x^2-18)(x+2)

Factor out 2 from the first one:

2(x^2-9)(x+2)

Factor again:

2(x+3)(x-3)(x+2)

This is good because we have three x variables now and because the starting contained x^3 means we should get three as well.

Answer 2

4m^2 - 28mn + 49n^2 {I'm guessing what you meant}

ac = 4(49) = 196

Factors of 196 that add to - 28 are -14 & -14

4m^2 - 14mn - 14mn + 49n^2 {rewrite for grouping}

2m(2m - 7n) - 7n(2m - 7n)

(2m - 7n)(2m - 7n)

2x^3 +4x^2 - 18x - 36 {already set for grouping}

2x^2(x + 2) - 18(x - 2)

(2x^2 - 18)(x + 2)

2(x^2 - 9)(x + 2)

2(x + 3)(x - 3)(x + 2)

Answer 3

4m^2-28mn+49n^2

(2m-7n)^2

2x^3+4x^2-18x-36

group here

2x^2(x+2)-18(x+2)

2x^2-18    x+2

2(x^2-9)   x+2

2(x-3)(x+3)  (x+2)

Answer 4

PROBLEM 1:

For the first one, you have coefficients that are squares (4 and 49) which should give you a clue that it is a perfect square.  Split the middle term into -14mn -14mn...

4m² - 14mn - 14mn + 49n²

Now factor out 2m from the first two terms:

2m(2m - 7n) - 14mn + 49n²

And factor out 7n from the last two terms:

2m(2m - 7n) - 7n(2m - 7n)

Finally factor out the common 2m - 7n:

(2m - 7n)(2m - 7n)

Look, it is a perfect square!

(2m - 7n)²

PROBLEM 2:

Pull out 2x² from the first two terms:

2x²(x + 2) - 18x - 36

Pull out a common 18 from the last two terms:

2x²(x + 2) - 18(x + 2)

Then the common x + 2:

(2x² - 18)(x + 2)

Now a common 2 from the first set:

2(x² - 9)(x + 2)

And then you have a difference of squares:

2(x - 3)(x + 3)(x + 2)

Answer 5

In the first one, both 4 and 49 are squares. I expect the variables are to the power of 2, so the factors are (2m-7n)(2m-7n)

In the second one, try to find a factor by trial and error. First, factor out the constant 2, giving x³+2x²-9x-18

Now the remaining constants can be seen to be double the previous ones (1 and 2, and 9 and 18). Try out (x+2) as a factor. If you divide, you get (2)(x+2)(x²-9). Factor that final one further so you get 2(x+2)(x-3)(x+3).

Answer 6

4m^2 - 28mn +49n^2 = (2m-7n)^2

2x^3 +4x^2 - 18x - 36= 2x^2(x+2)-18(x+2)=(x+2)(2x^2-18)

=(x+2)[2(x^2-9)]=2(x+2)(x-3)(x+3)