How do you factor completely $12 y^{2} + 9 y + 8 y + 6$ $12y^2+9y+8y+6$ ?

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How do you factor completely $12 y^{2} + 9 y + 8 y + 6$ $12y^2+9y+8y+6$ ?

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Answer 1 · 2015-11-19T23:44:31

Factor by grouping to find:

$12 y^{2} + 9 y + 8 y + 6 = (3 y + 2) (4 y + 3)$ $12y^2+9y+8y+6 = (3y+2)(4y+3)$

Explanation:

$12 y^{2} + 9 y + 8 y + 6$ $12y^2+9y+8y+6$

$= (12 y^{2} + 9 y) + (8 y + 6)$ $=(12y^2+9y)+(8y+6)$

$= 3 y (4 y + 3) + 2 (4 y + 3)$ $=3y(4y+3)+2(4y+3)$

$= (3 y + 2) (4 y + 3)$ $= (3y+2)(4y+3)$

Note that the hard work / fun has already been done for you in the split $9 y + 8 y$ $9y+8y$ . Normally you would be given a quadratic like $12 y^{2} + 17 y + 6$ $12y^2+17y+6$ to factor and have to find the split yourself.

If you are given something like $12 y^{2} + 17 y + 6$ $12y^2+17y+6$ to factor, then one way is to multiply $12 \times 6 = 72$ $12 xx 6 = 72$ then try to find a pair of factors that multiply to give $72$ $72$ and sum to give $17$ $17$ . Once you find the pair $9$ $9$ , $8$ $8$ works then you can split $17 y = 9 y + 8 y$ $17y=9y+8y$ and factor by grouping as we have done here.

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How do you factor completely $12 y^{2} + 9 y + 8 y + 6$ $12y^2+9y+8y+6$ ?

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How do you factor completely $12 y^{2} + 9 y + 8 y + 6$ $12y^2+9y+8y+6$ ?