Question

factor :8y^2+10y-3

Answer 1

To factor the quadratic expression \(8y^2 + 10y - 3\), we can use the method of factoring by grouping or applying the quadratic formula to find the roots.

The standard form of the quadratic is \(ax^2 + bx + c\), where:

- \(a = 8\)
- \(b = 10\)
- \(c = -3\)

We first calculate the product \(ac\):

\[
ac = 8 \times -3 = -24
\]

We need to find two numbers that multiply to \(-24\) (the value of \(ac\)) and add up to \(10\) (the value of \(b\)). The numbers that satisfy this condition are \(12\) and \(-2\) since:

\[
12 \times -2 = -24 \quad \text{and} \quad 12 + (-2) = 10
\]

Next, we can rewrite the middle term using these two numbers:

\[
8y^2 + 12y - 2y - 3
\]

Now we group the terms:

\[
(8y^2 + 12y) + (-2y - 3)
\]

Factor out the common factors in each group:

\[
4y(2y + 3) - 1(2y + 3)
\]

Now we can factor out the common binomial factor \((2y + 3)\):

\[
(2y + 3)(4y - 1)
\]

Thus, the factored form of the quadratic expression \(8y^2 + 10y - 3\) is:

\[
\boxed{(2y + 3)(4y - 1)}
\]