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How do you factor #10a^2 - 14a - 12#?

Answer 1

Kennedy Adams

#10a^2-14a-12=2(5a^2-7a-6)#

Use AC Method to find factors of #5a^2-7a-6# ...

#A=5#, #B=7#, #C=6#

Look for a pair of factors of #AC=5*6=30# whose difference is #B=7#.

The pair #B1=10#, #B2=3# works.

Then for each of the pairs #(A, B1)#, #(A, B2)# divide by the HCF to get the coefficients of a factor, choosing signs as appropriate:

#(A, B1) = (5, 10) -> (1, 2) -> (a-2)# #(A, B2) = (5, 3) -> (5, 3) -> (5a+3)#

So:

#10a^2-14a-12=2(5a^2-7a-6) = 2(a-2)(5a+3)#

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Answer 2

Nathan Axel

To factor 10a^2 - 14a - 12, you first find the product of the coefficient of the squared term (10) and the constant term (-12), which equals -120. Then, find two numbers that multiply to -120 and add to the coefficient of the linear term (-14). These numbers are -20 and 6. Rewrite the middle term using these numbers: 10a^2 - 20a + 6a - 12. Factor by grouping: 2a(5a - 10) + 6(5a - 10). Finally, factor out the common factor (5a - 10): (2a + 6)(5a - 10).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.