6y^2-48+42y...factor compleatly?

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6y^2-48+42y...factor compleatly?

2007-06-28 13:58:22 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

4 answers

6y^2 - 48 + 42y
= 6y^2 + 42y - 48
= 6(y^2 + 7y - 8)
= 6(y-1)(y+8)

2007-06-28 14:02:33 · answer #1 · answered by Kemmy 6 · 0⤊ 0⤋

6y^2 - 48 + 42y

= 6y^2 + 42y - 48 write in standard form

= 6(y^2 + 7y - 8) factor out greatest common factors

= 6[ y^2 - 1y + 8y - 8 ] substitute sum of factors of 8y^2 in place of 7y

= 6[ y(y - 1) + 8(y - 1) ] factor by grouping

= 6[ (y - 1)(y + 8) ] factor out common binomial factor

Answer: 6(y - 1)(y + 8)

2007-06-28 21:14:24 · answer #2 · answered by mathjoe 3 · 0⤊ 0⤋

= 6.(yÂ² + 7y - 8)
= 6.(y + 8).(y - 1)

2007-07-02 12:06:12 · answer #3 · answered by Como 7 · 0⤊ 0⤋

no it is not

Try FOIL

2007-06-28 21:02:06 · answer #4 · answered by Eric S 1 · 0⤊ 1⤋