# How do I factor this math problem?

I'm stuck on this.

20w^2+13w-15

20w^2 + 13w - 15

(5w - 3)(4w + 5)

(This is 20w^2 + 25w - 12w - 15)

#1

20w^2 + 13w - 15

(5w - 3)(4w + 5)

#2

20w^2 + 25w -12w -15

5w ( 4w +5 ) -3 ( 4w +5)

(5w - 3 ) ( 4w + 5 )

#3

(5w-3)(4w+5)

for questions like this look at factors of 20 i.e. 2x10 4x5

and -15 i.e. -3x5 -5x3

then just use trial and error with different cominations until you get one that

can yield 13w when you factor

#4

Let Q = 20w^2 + 13w - 15

then 20Q = 400w^2 + 13(20w) - 300

20Q = (20w)^2 + 13(20w ) - 300

Let u = 20w

20Q = u^2 + 13u - 300

20Q = (u + 25 )(u - 12 )

20Q = (20w + 25 )(20w - 12 )

20Q = 5(4w + 5 )*4(5w - 3 )

Q = (5w - 3 )(4w + 5 )

#5

The prime factorization of 20: 1, 2, 4, 5, 10, 20

The prime factorization of 15: 1, 3, 5, 15

The two numbers that multiply together to get 20 will both be positive.

For the two numbers that multiply together to get -15, one will be positive, the other will be negative.

This is the layout for factoring a trinomial: (a + b)(c + d)

a×c = 20w2

b×d = -15

ad + bc = 13w

Since bc is negative and the final result is positive, you know that ad must have a greater absolute value than bc.

Now, you mix and match with the factors until you get a result.

...wow, I can't even do it on my own. This is like trying to solve a sudoku puzzle. Now, you can use a factoring calculator: http://www.solvemymath.com/online_math_c…

the answer is (4w + 5)(5w - 3).

#6