Q 118.

Factor Trinomials of the Form $a x^{2} + b x + c$ using the ‘ac’ Method.

$12 z^{2} - 41 z - 11$

Verified

Answer

The solution is $\begin{array}{rcl} (4 z + 1) (3 z - 11) \end{array}$ .

1Step 1. Given information

The given trinomial is $12 z^{2} - 41 z - 11$ .

2Step 2. Find greatest common factor.

There is no greatest common factor.

3Step 3. Compare the given trinomial to a x 2 + b x + c and then find the product of a c .

The values are as follows:

$\begin{array}{rcl} a & = & 12 \\ b & = & - 41 \\ c & = & - 11 \end{array}$

The product is as follows:

$\begin{array}{rcl} a c & = & 12 (- 11) \\ = & - 132 \end{array}$

4Step 4. Determine two numbers that multiply to a c and add to b .

Two numbers that multiply to $a c$ and add to $b$ are as follows:

$\begin{array}{rcl} 3 (- 44) & = & - 132 \\ = & a c \\ 3 - 44 & = & - 41 \\ = & b \end{array}$

So, two numbers are $3$ and $- 44$ .

Now, split the middle term of the given trinomial $12 z^{2} - 41 z - 11$ as follows:

$12 z^{2} - 41 z - 11 = 12 z^{2} + 3 z - 44 z - 11$

5Step 5. Factor by grouping.

$\begin{array}{rcl} 12 z^{2} - 41 z - 11 & = & 3 z (4 z + 1) - 11 (4 z + 1) \\ = & (4 z + 1) (3 z - 11) \end{array}$

6Step 6. Check by multiplying ( 4 z + 1 ) ( 3 z - 11 ) .

$\begin{array}{rcl} (4 z + 1) (3 z - 11) & = & 12 z^{2} + 3 z - 44 z - 11 \\ = & 12 z^{2} + 41 z - 11 \end{array}$

Hence, the factor is $\begin{array}{rcl} (4 z + 1) (3 z - 11) \end{array}$ .