Does 4y²+7y-2 factor?

(4y - 1)(y + 2)

:)

(4y - 1)(y + 2)

4y^2 + 7y - 2
= (4y - 1)(y + 2)

4y² + 7y - 2 = (4y + b)(y - d) = 4y² + (b-4d)y - bd. Choosing b = -1 and d = -2 yields 4y² + (-1+8)y - (-1)(-2) = y² + 7y - 2.

So 4y² + 7y - 2 = (4y - 1)(y + 2)

instead of just saying "yes" and giving you the factors, here's the work:

4y²+7y-2
multiply 4 and 2, and find factors of that answer that add or subtract to 7
4*2=8
8-1=7
now expand that 7 into 8-1 and then factor twice

4y²+7y-2
4y²+(8-1)y-2
4y²+8y-1y-2
4y(y+2) - 1(y+2) <--factor once
(4y-1)(y+2) <--factor twice <--- final answer

yes, it is factor...
factorise the equation..u will get, (4y -1)(y +2)...

y(4y+7-2/y)

The discriminant is 7^2 - 4*4*-2 = 81
This is a perfect square so it does factor.

(4y-1)(y+2)

4y^2 + 7y - 2
= (4y - 1)(y + 2)