Question

Factorise $a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}$.

Answer 1

- Hint: Here, factorisation of $a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}$ means we have to decompose it into product of factors. First take the common factor outside from $a{{b}^{2}}-ab$ and $-{{a}^{2}}b+{{a}^{2}}$. Then we have to take outside the common factor from the above two terms which is $b-1$. At the end we can write $a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}$ as the product of three factors.

Complete step-by-step solution -

We are given $a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}$. Now we have to factorise the given equation.
We know that factorisation is defined as breaking or decomposition of a polynomial into a product of factors which when multiplied together give the original polynomial.
 First consider, $a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}$.
Consider the first two terms together and the last two terms together.
For the first two terms $a{{b}^{2}}$ and $-ab$, ab is a common factor, so we can take outside ab.
Then we will get the equation:
$a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}=ab(b-1)-{{a}^{2}}b+{{a}^{2}}$
Next, from the last two terms $-{{a}^{2}}b$ and ${{a}^{2}}$, ${{a}^{2}}$ is a common factor. So you can take outside ${{a}^{2}}$, we will get:
$a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}=ab(b-1)+{{a}^{2}}(-b+1)$
 Now, in the first term you have $b-1$ and in the second term you have $-b+1$ where:
$-b+1=-(b-1)$
 Therefore, you can take outside -1 from $-b+1$ . Hence, we will get:
$a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}=ab(b-1)-{{a}^{2}}(b-1)$
 Now, we can say that $b-1$ is common factor for the two terms. So, take outside $b-1$, we will get:
 $a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}=(b-1)(ab-{{a}^{2}})$
 Next, from $ab-{{a}^{2}}$, $a$ is a common factor, so you can take outside $a$.
Now we will get the equation:
$a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}=(b-1)a(b-a)$
 Hence, we can write it as:
$a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}=a(b-1)(b-a)$
 Hence, $a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}$ has the factors $a,b-1$ and $b-a$ and we can write $a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}$ as product of these 3 factors.

Note: Here, we can also factorise $a{{b}^{2}}-ab-{{a}^{2}}b+{{a}^{2}}$ by rearranging it as $a{{b}^{2}}-{{a}^{2}}b-ab+{{a}^{2}}$ or
$a{{b}^{2}}+{{a}^{2}}-{{a}^{2}}b-ab$. Now consider the first two terms together and the last two terms together and find the common factors and factorise. You will get the same factors $a,b-1$ and $b-a$ at the end.