7(x^2)+33x-10
This is the way I would approach this problem.
Find the factors of the first and last number:
Factors of 7: 1,7
Factors of -10: -10,-5,-2,-1,1,2,5,10
Set up the two quantities:
(x +/- __)(7x +/- __ )
Figure out which factors of -10 when multiplied with the factors of 7 then added together will equal +33:
5(7)+(-2)=33
Therefore:
(x+5)(7x-2)
Double check answer by foiling:
(x+5)(7x-2)=7(x^2)-2x+35x-10
7(x^2)+33x-10
7x2 + 68xy - 20y2
(x + 5)(x^2 + 2x + 2)
7x2 plus 7x5 = 49
It can't be factored because its discriminant is less than zero
2x+5x-24 7x2-24
7x2 + 68xy - 20y2
(7x2 + 9)(4x - 1)
The GCF is x2
7(x-5)^2
(x + 5)(x^2 + 2x + 2)
7x2 + 7x - 14 = 7(x2 + x - 2) = 7(x + 2)(x - 1)
7x2
7x2 plus 7x5 = 49
It can't be factored because its discriminant is less than zero
2x+5x-24 7x2-24
You can take 7x2 out of both of those, leaving 5x4 and 1.
7(x2 + 7)