Question

completely factor the trinomial, if possible.

\\2r^2 + 21r + 10\\

Explanation:

Identify the trinomial coefficients

We want to factor the quadratic trinomial:

$$ 2r^2 + 21r + 10 $$

Using the Factoring Trinomials knowledge point, we identify the coefficients:

• \(a = 2\)
• \(b = 21\)
• \(c = 10\)

Find factors for grouping

We look for two numbers that multiply to \(a \cdot c\) and add to \(b\):

• Product: \(2 \cdot 10 = 20\)
• Sum: \(21\)

The two numbers that satisfy these conditions are \(20\) and \(1\), since:

$$ 20 \cdot 1 = 20 \quad \text{and} \quad 20 + 1 = 21 $$

Rewrite and group terms

We rewrite the middle term \(21r\) using \(20r\) and \(1r\):

$$ 2r^2 + 20r + r + 10 $$

Now, group the terms into two pairs:

$$ (2r^2 + 20r) + (r + 10) $$

Factor out the GCF from each group

Factor out the greatest common factor from each group:

• From \(2r^2 + 20r\), factor out \(2r\):

$$ 2r(r + 10) $$

• From \(r + 10\), factor out \(1\):

$$ 1(r + 10) $$

Combining these gives:

$$ 2r(r + 10) + 1(r + 10) $$

Factor out the common binomial

Factor out the common binomial factor \((r + 10)\):

$$ (2r + 1)(r + 10) $$

Answer:

\((2r + 1)(r + 10)\)