Question

difficulty: ★★★
factorise completely $3h^2 + 11h + 6$
a $(h + 1)(3h + 6)$
b $(h + 6)(3h + 1)$
c $(h + 3)(3h + 2)$
d $(h + 2)(3h + 3)$

Explanation:

Step1: Recall factoring quadratic formula

For a quadratic \(ax^2 + bx + c\), we find two numbers that multiply to \(a\times c\) and add to \(b\). Here, \(a = 3\), \(b = 11\), \(c = 6\). So \(a\times c=3\times6 = 18\). We need two numbers that multiply to \(18\) and add to \(11\). The numbers are \(9\) and \(2\) (since \(9\times2 = 18\) and \(9 + 2=11\)).

Step2: Split the middle term

Rewrite \(11h\) as \(9h+2h\):
\(3h^2 + 9h+2h + 6\)

Step3: Group and factor

Group the first two and last two terms:
\((3h^2 + 9h)+(2h + 6)\)
Factor out common factors from each group:
\(3h(h + 3)+2(h + 3)\)

Step4: Factor out the common binomial

Now, factor out \((h + 3)\):
\((h + 3)(3h + 2)\)

We can also check by expanding each option:

• Option A: \((h + 1)(3h + 6)=3h^2+6h+3h + 6=3h^2+9h + 6\) (not equal to original)
• Option B: \((h + 6)(3h + 1)=3h^2+h+18h + 6=3h^2+19h + 6\) (not equal)
• Option C: \((h + 3)(3h + 2)=3h^2+2h+9h + 6=3h^2+11h + 6\) (matches)
• Option D: \((h + 2)(3h + 3)=3h^2+3h+6h + 6=3h^2+9h + 6\) (not equal)

Answer:

C. \((h + 3)(3h + 2)\)