Q 113.

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

Verified

Answer

The solution is $(2 k - 3) (2 k - 5)$ .

1Step 1. Given information

The given trinomial is $4 k^{2} - 16 k + 15$ .

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:

$\begin{array}{rcl} a & = & 4 \\ b & = & - 16 \\ c & = & 15 \end{array}$

The product is as follows:

$\begin{array}{rcl} a c & = & 4 \cdot 15 \\ = & 60 \end{array}$

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

Since the middle term in the given polynomial is negative, both factors will be negative.

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

$\begin{array}{rcl} (- 6) (- 10) & = & 60 \\ = & a c \\ - 6 - 10 & = & - 16 \\ = & b \end{array}$

So, two numbers are $- 6$ and $- 10$ .

Now, split the middle term of the given trinomial $4 k^{2} - 16 k + 15$ as follows:

width="251" style="max-width: none; vertical-align: -4px;" $4 k^{2} - 16 k + 15 = 4 k^{2} - 6 k - 10 k + 15$

5Step 5. Factor by grouping.

$\begin{array}{rcl} 4 k^{2} - 16 k + 15 & = & 2 k (2 k - 3) - 5 (2 k - 3) \\ = & (2 k - 3) (2 k - 5) \end{array}$

6Step 6. Check by multiplying ( 2 k - 3 ) ( 2 k - 5 ) .

$\begin{array}{rcl} (2 k - 3) (2 k - 5) & = & 4 k^{2} - 10 k - 6 k + 15 \\ = & 4 k^{2} - 16 k + 15 \end{array}$

Hence, the factor is $\begin{array}{rcl} (2 k - 3) (2 k - 5) \end{array}$ .