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Chapter 6: Q. 373 (page 635)

the following exercise, factor.

18a2鈭�9a+1

Short Answer

Expert verified

The factorisation of the given polynomial is:

18a2鈭�9a+1=(3a-1)(6a-1)

Step by step solution

01

Step 1. Given information

The given expression is18a2鈭�9a+1

02

Step 2. Use ac method for factorisation

To factorise the polynomial ax2+bx+cby ac method, we need to think of two numbers whose product is equal to acand the sum is equal to b.

For polynomial 18a2-9a+1we have:

ac=18脳1=18

And we have to think of two numbers whose sum is equal to 9and product is equal to localid="1644609316381" 18.

Then

-6-3=9-6脳-3=9

So the required numbers arelocalid="1644608815338" -6&-3

03

Step 3. Perform factorisation

Now,

18a2鈭�9a+118a2-6a-3a+16a(3a-1)-(3a-1)[Takingout6aand-1asacommon](6a-1)(3a-1)[Takingout3a-1asacommon]

Thus, factorisation of given polynomial is:

18a2鈭�9a+1=(3a-1)(6a-1)

04

Step 4. Check the answer

Multiplying the factors, we get:

18a2-9a+1=3a-1(6a-1)18a2-9a+1=18a2-6a-3a+118a2-9a+1=18a2-9a+1

This is true.

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