How do you factor #(4y-5)^2+3(4y-5)-70#?

1 Answer
Noah G
Jun 13, 2017

This can be factored as #4(4y +5)(y - 3)#

Explanation:

We let #a = 4y - 5#. Then the polynomial can be rewritten as:

#=a^2 + 3a - 70#

Which is a standard polynomial of the form #ax^2 + bx + c#. We find two numbers that multiply to #c# and that add to #b#. These would be #+10# and #-7#.

#=(a + 10)(a - 7)#

Resubstitute to get:

#=(4y - 5 + 10)(4y - 5 - 7)#

#= (4y + 5)(4y - 12)#

#=4(4y + 5)(y - 3)#

Hopefully this helps!

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