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

Jul 20, 2015

$\left(5 d - 4\right) \left(2 d + 5\right)$

Explanation:

We are looking for a solution of the form:
$\left(a d + b\right) \left(e d + f\right) = \left(a e\right) {d}^{2} + \left(a f + e b\right) d + b f$

So we need to solve the simultaneous equations:
$a e = 10$
$a f + e b = 17$
$b f = - 20$

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:
$10 {d}^{2} + 17 d - 20 = \left(5 d - 4\right) \left(2 d + 5\right)$

Jul 20, 2015

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)

Jul 21, 2015

$10 {d}^{2} + 17 d - 20 = \left(2 d + 5\right) \left(5 d - 4\right)$

Explanation:

$10 {d}^{2} + 17 d - 20$ is a quadratic equation in the form $a {x}^{2} + b x + c$, where $a = 10 , b = 17 , \mathmr{and} c = - 20$.

Factor by grouping, also called the $a \cdot c$ method of factoring, and factoring by splitting the middle term.

Multiply $a \cdot c$

$10 \cdot - 20 = - 200$

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 $25 d \mathmr{and} - 8 d$ for $17 d$.

$10 {d}^{2} + 25 d - 8 d - 20$

Group the terms into two groups.

$\left(10 {d}^{2} + 25 d\right) - \left(8 d - 20\right)$

Factor out the GCF for each group of terms.

$5 d \left(2 d + 5\right) - 4 \left(2 d + 5\right)$

Factor out the common term.

$\left(2 d + 5\right) \left(5 d - 4\right)$