Question

How do you factor `10d^2 + 17d -20`?

Answer 1

`(5d-4)(2d+5)`

Explanation:

We are looking for a solution of the form:
`(ad+b)(ed+f)=(ae)d^2+(af+eb)d+bf`

So we need to solve the simultaneous equations:
`ae=10`
`af+eb=17`
`bf=-20`

This has a solution (not unique - this solution is chosen as all terms are integers):
`a=5,b=-4,e=2,f=5`

We then have:
`10d^2+17d-20=(5d-4)(2d+5)`

Answer 2

Factor: y = 10 x^2 + 17x - 20
Answer: y = (5x - 4)(2x + 5)

Explanation:

I use the new AC Method to factor trinomials (Google, Yahoo Search).
y = 10x^2 + 17x - 20 = 10(x - p)(x - q)
Converted y' = x^2 + 10x - 200.= (x - p')(x - q'). p' and q' have opposite signs.
Factor pairs of (-200) -> (-4, 50)(-8, 25). This sum is 17 = b.
Then p' = -8, and q' = 25.
Then, p = (p')/a = -8/10 = -4/5, and q' = 25/10 = 5/2.
Factored form: y = 10(x - 4/5)(x + 5/2) = (5x - 4)(2x + 5)

Answer 3

`10d^2+17d-20=(2d+5)(5d-4)`

Explanation:

`10d^2+17d-20` is a quadratic equation in the form `ax^2+bx+c`, where `a=10, b=17, and c=-20`.

Factor by grouping, also called the `a*c` method of factoring, and factoring by splitting the middle term.

Multiply `a*c`

`10*-20=-200`

Find two numbers that when added equal `17`, and when multiplied equal `-200`.

The numbers `25` and `-8` satisfy the requirements.

Rewrite the equation substituting the sum of `25d and -8d` for `17d`.

`10d^2+25d-8d-20`

Group the terms into two groups.

`(10d^2+25d)-(8d-20)`

Factor out the GCF for each group of terms.

`5d(2d+5)-4(2d+5)`

Factor out the common term.

`(2d+5)(5d-4)`