Q. 373 - Intermediate Algebra

Question

the following exercise, factor.

$18a^2 - 9a + 1$

Step-by-Step Solution

Answer

The factorisation of the given polynomial is: 

$18a^2 - 9a + 1=(3a-1)(6a-1)$

Step 1. Given information

The given expression is $18a^2- 9a + 1$

Step 2. Use ac method for factorisation

To factorise the polynomial $a{x}^2+bx+c$ by ac method, we need to think of two numbers whose product is equal to $ac$ and the sum is equal to  $b$.

For polynomial $18a^2-9a+1$we have:

$$\begin{gathered} ac=18\times 1 \\ =18 \end{gathered}$$

And we have to think of two numbers whose sum is equal to $9$ and product is equal to $18$.

Then 

$$\begin{gathered} -6-3=9 \\ -6\times -3=9 \end{gathered}$$ 

So the required numbers are $-6 \& -3$

Step 3. Perform factorisation

Now,

$$\begin{gathered} 18a^2 - 9a + 1 \\ 18a^2-6a-3a+1 \\ 6a(3a-1)-(3a-1) [Taking out 6a and -1 as a common] \\ (6a-1)(3a-1) [Taking out 3a-1 as a common] \end{gathered}$$

Thus, factorisation of given polynomial is:

$18a^2 - 9a + 1=(3a-1)(6a-1)$

Step 4. Check the answer

Multiplying the factors, we get:

$$\begin{gathered} 18a^2-9a+1=\left(3a-1\right)(6a-1) \\ 18a^2-9a+1=18a^2-6a-3a+1 \\ 18a^2-9a+1=18a^2-9a+1 \end{gathered}$$

This is true.