How do you factor #7x^2+5x-2#?

Question

6347 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-24T15:04:24

#(7x-2)(x+1)#

#7x^2# only has #7x and x# as factors
-2 is #2xx1# with one of them negative.

Answer 2 · 2018-04-24T15:09:20

#(x+1)(7x-2)#

Multiply the coefficient by the constant:

#7 xx -2=-14#

Replace it as the new constant:

#-> 7x^2+5x-14#

Find the numbers:

#7-2=5#

#7xx-2=-14#

So these work.

Substitute these in:

#7x^2+7x-2x-2# #color(blue)(lArr "Notice we replaced the 14 with 2")#

Factorise the first two expressions:

#7x^2+7x -> 7x(x+1)#

Factorise the next two expressions:

#-2x-2#

#-2(x+1)# #color(blue)(lArr "Notice the brackets are the same")#

Therefore we get:

#color(red)(7x)color(green)((x+1))color(red)(-2)color(green)((x+1))#

One bracket is #(x+1)# and the other bracket is the remaining terms:

#-> (7x-2)#

We can always check by expanding:

#(7x-2)(x+1)=7x^2+7x-2x-2#

#-> 7x^2+5x-2#

Which is the original, so we are correct.