Question

factor.
$20y^{2}-13y - 21$

Explanation:

Step1: Find two numbers for ac

For the quadratic \(20y^{2}-13y - 21\), \(a = 20\), \(b=-13\), \(c=-21\). Calculate \(ac=20\times(-21)=- 420\). We need two numbers that multiply to \(-420\) and add to \(-13\). The numbers are \(-28\) and \(15\) since \(-28\times15=-420\) and \(-28 + 15=-13\).

Step2: Rewrite the middle term

Rewrite \(-13y\) as \(-28y+15y\): \(20y^{2}-28y + 15y-21\).

Step3: Group and factor

Group the first two and last two terms: \((20y^{2}-28y)+(15y - 21)\). Factor out \(4y\) from the first group and \(3\) from the second group: \(4y(5y - 7)+3(5y - 7)\).

Step4: Factor out the common binomial

Factor out \((5y - 7)\): \((4y + 3)(5y - 7)\).

Answer:

\((4y + 3)(5y - 7)\)