How do you factor #7x^2 - 9x - 10#?

2 Answers
Jun 22, 2016

Answer:

#(7x+5)(x-2)#

Explanation:

Let's start by finding the zeroes of the trynomial:

#x=(9+-sqrt(81-4(7)(-10)))/(2(7))#

#x=(9+-sqrt(81+280))/4#

#x=(9+-sqrt(361))/14#

#x=(9+-19)/14#

#x=-10/14 and x=28/14#

that's

#x_1=-5/7 and x_2=2#

Since a trynomial can be factorized by the formula:

#ax^2+bx+c=a(x-x_1)(x-x_2)#

so

#7x^2-9x+10=7(x+5/7)(x-2)#

or

#(7x+5)(x-2)#

Jun 23, 2016

Answer:

(7x + 5)(x - 2)

Explanation:

Use the new AC method to factor trinomials (Socratic Search)
#y = 7x^2 - 9x - 10 =# 7(x + p)(x + q)
Converted trinomial: #y' = x^2 - 9x - 70 =# (x + p')(x + q').
Find p' ans q' knowing they have opposite sign since ac < 0.
Factor pairs of (ac = -70) --> ...(-5, 14)(5, - 14). This sum is (-9 = b). Then, p' = 5 and q' = -14.
Back to original y --> #p = (p')/a = 5/7#, and #q = (q')/a = -14/7 = -2#.
Factored form: #y = 7(x + 5/7)(x - 2) = (7x + 5)(x - 2)#