How do you factor #10x^2+15x-70#?

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Answer 1 · 2017-10-22T15:36:46

The correct factoring is #5(2x + 7)(x - 2)#

We can start getting rid of a #5#:

#5(2x^2 + 3x - 14)#

Now we rewrite as

#5(2x^2 - 4x + 7x - 14)#

Which is equivalent as #-4x + 7x = 3x#. Now notice that we can factor further:

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

We can now factor out #(x - 2)#.

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

If we try expanding this we see

#5(2x^2 + 7x - 4x - 14)#

#10x^2 + 15x - 70#

Which is what we had at first, so the factoring is correct.

Hopefully this helps!

Answer 2 · 2017-10-22T15:45:42

$$10x^2+15x-70=5(2x+7)(x-2)$$

Take out the common factor of 5 to give: $$5(2x^2+3x-14)$$
You know that the brackets will next look like this:
$$5(2x ± ??)(x ± ??)$$
You need to determine what two numbers when multiplied will equal -14 and will fit in with the 2x. A quick bit of trial and error shows that the numbers that will fit are +7 and -2 which gives: $$5(2x+7)(x-2)$$