How do you factor $4y^2 + 25y + 6$ ?

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Answer 1 · 2017-03-03T09:59:34

Factors of given qudratic equation are : $(y+6)(4y+1)$

$4y^2+25y+6$

$4y^2+24y+y+6$

$ul(4y^2+24y)+ul(y+6)$

$4y(y+6)+(y+6)$

Taking common $y+6$

Factors : $(y+6)(4y+1)$

How do you factor 4y^2 + 25y + 64y^2 + 25y + 6?