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

3 Answers
Jul 20, 2015

#(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)#

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

#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)#