Question

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

Explanation:

Step1: Multiply \(a\) and \(c\)

For quadratic \(ax^2 + bx + c\), here \(a = 3\), \(b=-11\), \(c = 6\).
\(a\times c=3\times6 = 18\)

Step2: Find factors of 18

Find two numbers that multiply to \(18\) and add to \(b=-11\).
The numbers are \(-9\) and \(-2\) (since \(-9\times -2 = 18\) and \(-9 + (-2)=-11\))

Step3: Split the middle term

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

Step4: Group and factor

Group first two and last two terms:
\((3x^2 - 9x)+(-2x + 6)\)
Factor out common terms:
\(3x(x - 3)-2(x - 3)\)
Factor out \((x - 3)\):
\((x - 3)(3x - 2)\)

Answer:

A. \((x - 3)(3x - 2)\)