Question

which of the following shows $4b^2 - 12b + 9$ factored completely?
a
$(2b + 3)^2$

b.
$(2b - 9)(2b - 1)$

c.
$(2b - 3)^2$

d.
$(2b - 3)(2b + 3)$

Explanation:

Step1: Recall perfect square formula

The perfect square trinomial formula is \(a^2 - 2ab + b^2=(a - b)^2\) and \(a^2+2ab + b^2=(a + b)^2\). For the quadratic \(4b^2-12b + 9\), we can rewrite \(4b^2=(2b)^2\), \(9 = 3^2\), and \(- 12b=-2\times(2b)\times3\).

Step2: Apply the formula

Comparing with \(a^2-2ab + b^2\) where \(a = 2b\) and \(b = 3\) (in the formula, the second \(b\) here is a constant, not the variable \(b\) in the quadratic, to avoid confusion we can use \(c\) instead, so \(a^2-2ac + c^2=(a - c)^2\) with \(a = 2b\), \(c = 3\)), we get \(4b^2-12b + 9=(2b-3)^2\).

We can also check other options:

• Option A: \((2b + 3)^2=(2b)^2+2\times(2b)\times3+3^2 = 4b^2 + 12b+9

eq4b^2-12b + 9\)

• Option B: \((2b-9)(2b - 1)=4b^2-2b-18b + 9=4b^2-20b + 9

eq4b^2-12b + 9\)

• Option D: \((2b-3)(2b + 3)=(2b)^2-3^2=4b^2-9

eq4b^2-12b + 9\)

Answer:

C. \((2b - 3)^2\)