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How do you factor #12y^(2)-4y-5#?

Answer 1

Eli Assouline

#12y^2-4y-5 = (2y+1)(6y-5)#

Use an AC method:

Given:

#12y^2-4y-5#

Look for a pair of factors of #AC = 12*5 = 60# which differ by #B=4#.

The pair #10, 6# works.

Use this pair to split the middle term and factor by grouping:

#12y^2-4y-5 = (12y^2-10y)+(6y-5)#

#color(white)(12y^2-4y-5) = 2y(6y-5)+1(6y-5)#

#color(white)(12y^2-4y-5) = (2y+1)(6y-5)#

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Answer 2

Ethan Adkins

To factor the quadratic expression (12y^2 - 4y - 5), you can use the factoring method. First, determine if the quadratic can be factored using the product-sum method. If not, you can use the quadratic formula. In this case, the quadratic can be factored using the product-sum method. Find two numbers that multiply to give (12 \times (-5) = -60) and add to give the coefficient of the middle term, which is -4. The numbers are -10 and 6. Rewrite the middle term using these numbers: (12y^2 - 10y + 6y - 5). Factor by grouping: (2y(6y - 5) + 1(6y - 5)). Factor out the common factor: ((2y + 1)(6y - 5)). So, the factored form of (12y^2 - 4y - 5) is ((2y + 1)(6y - 5)).

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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.