How do you factor #5x^2-17x+14#?

2 Answers
Jan 25, 2017

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

Explanation:

First you would find the zeros #x_1,x_2# of the polynomial #5x^2-17x+14#

You would apply the quadratic formula:

#x_(1,2)=(17+-sqrt(17^2-4*5*14))/(2*5)#

#=(17+-sqrt(289-280))/10=(17+-3)/10#

#x_1=14/10=7/5# and #x_2=20/10=2#

Then you would factor the given polynomial by applyng the relation:

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

So you get:

#5x^2-17x+14=5(x-7/5)(x-2)=(5x-7)(x-2)#

Jan 25, 2017

You can use an AC method to find:

#5x^2-17x+14 = (5x-7)(x-2)#

Explanation:

This quadratic can be factored using an AC method:

Find a pair of factors of #AC = 5*15 = 70# with sum #B=17#

The pair #10, 7# works.

Use this pair to split the middle term and factor by grouping:

#5x^2-17x+14 = 5x^2-10x-7x+14#

#color(white)(5x^2-17x+14) = (5x^2-10x)-(7x-14)#

#color(white)(5x^2-17x+14) = 5x(x-2)-7(x-2)#

#color(white)(5x^2-17x+14) = (5x-7)(x-2)#