Question

Factor completely.
8t2–14t+3

Answer 1

To factor the quadratic polynomial \(8t^2 - 14t + 3\), we will use the method of factoring by grouping.

First, we need to identify two numbers that multiply to the product of the coefficient of \(t^2\) (which is 8) and the constant term (which is 3). The product is:

\[
8 \times 3 = 24
\]

Next, we need two numbers that multiply to 24 and add up to the coefficient of \(t\) (which is -14). The numbers that work are -12 and -2:

\[
-12 + (-2) = -14 \quad \text{and} \quad -12 \times -2 = 24
\]

Now we can rewrite the middle term \(-14t\) using -12 and -2:

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

Next, we group the terms:

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

Now, we can factor out the greatest common factor from each group:

\[
4t(2t - 3) - 1(2t - 3)
\]

Notice that \(2t - 3\) is common to both groups:

\[
(4t - 1)(2t - 3)
\]

Thus, the completely factored form of \(8t^2 - 14t + 3\) is:

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