Q 107.

Factor Trinomials of the Form $a x^{2} + b x + c$ Using Trial and Error.

$- 16 x^{2} - 32 x - 16$

Verified

Answer

The solution is $- 16 (x + 1) (x + 1)$ .

1Step 1. Given information

Consider the trinomial.

$- 16 x^{2} - 32 x - 16$

2Step 2. Write the trinomial in descending order and then find the greatest common factor.

The trinomial $- 16 x^{2} - 32 x - 16$ is given in descending order.

There is a greatest common factor in the given trinomial.

Factor the greatest common factor.

$- 16 x^{2} - 32 x - 16 = - 16 (x^{2} + 2 x + 1)$

3Step 3. Find the factors of the first term and the last term of the trinomial x 2 + 2 x + 1 .

The first term of the trinomial $x^{2} + 2 x + 1$ is $x^{2}$ .

So, the only factors of $x^{2}$ are as follows:

$x^{2} = x \cdot x$

The last term of the trinomial $x^{2} + 2 x + 1$ is $1$ .

Since all the terms in the polynomial $x^{2} + 2 x + 1$ is positive, all the factors of the last term will be positive.

The only factors of $1$ are as follows:

$1 = 1 \cdot 1$

4Step 4. Make a table for all the combination of factors of x 2 + 2 x + 1 .

Possible factors	Product
$(x + 1) (x + 1)$	$x^{2} + 2 x + 1$

From the table, conclude that the only combination is $(x + 1) (x + 1)$ .

So, the factor of the given trinomial is $- 16 (x + 1) (x + 1)$ .

5Step 5. Check by multiplying all the factors - 16 ( x + 1 ) ( x + 1 ) .

$\begin{array}{rcl} - 16 (x + 1) (x + 1) & = & - 16 (x^{2} + x + x + 1) \\ = & - 16 (x^{2} + 2 x + 1) \\ = & - 16 x^{2} - 32 x - 16 \end{array}$

Hence, the factor is $- 16 (x + 1) (x + 1)$ .