Question

factor completely. 2s² + 3s - 9

Explanation:

Step1: Check for common factors

The quadratic expression is \(2s^{2}+3s - 9\). There is no common factor among the three terms (2, 3, -9 have no common factor other than 1).

Step2: Use the AC method for factoring \(ax^{2}+bx + c\)

For \(2s^{2}+3s - 9\), where \(a = 2\), \(b=3\), \(c=- 9\). Calculate \(ac=2\times(-9)=-18\).
We need two numbers that multiply to -18 and add up to 3. The numbers are 6 and - 3 (since \(6\times(-3)=-18\) and \(6+( - 3)=3\)).

Step3: Rewrite the middle term

Rewrite \(3s\) as \(6s-3s\) in the expression:
\(2s^{2}+6s - 3s-9\)

Step4: Group the terms

Group the first two and the last two terms:
\((2s^{2}+6s)+(-3s - 9)\)

Step5: Factor out the common factor from each group

From the first group \(2s^{2}+6s\), factor out \(2s\): \(2s(s + 3)\)
From the second group \(-3s-9\), factor out - 3: \(-3(s + 3)\)
So the expression becomes \(2s(s + 3)-3(s + 3)\)

Step6: Factor out the common binomial factor

Factor out \((s + 3)\) from both terms:
\((2s-3)(s + 3)\)

Answer:

\((2s - 3)(s + 3)\)