Examples (Revised) - Chapter 14 - Factorisation - Ncert Solutions class 8 - Maths

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Chapter 12: Factorisation - NCERT Solutions for Class 8 Maths

Example 1:

Factorise
$
\begin{aligned}
& 12 a^2 b+15 a b^2 \\
& 12 a^2 b=2 \times 2 \times 3 \times a \times a \times b \\
& 15 a b^2=3 \times 5 \times a \times b \times b
\end{aligned}
$

Solution:

We have

The two terms have $3, a$ and $b$ as common factors.
Therefore,
$
\begin{aligned}
12 a^2 b+15 a b^2 & =(3 \times a \times b \times 2 \times 2 \times a)+(3 \times a \times b \times 5 \times b) \\
& =3 \times a \times b \times[(2 \times 2 \times a)+(5 \times b)] \\
& =3 a b \times(4 a+5 b) \\
& =3 a b(4 a+5 b) \quad \text { (required factor form) }
\end{aligned}
$
(combining the terms)

Example 2:

Factorise $10 x^2-18 x^3+14 x^4$
Solution:
$
\begin{aligned}
10 x^2 & =2 \times 5 \times x \times x \\
18 x^3 & =2 \times 3 \times 3 \times x \times x \times x \\
14 x^4 & =2 \times 7 \times x \times x \times x \times x
\end{aligned}
$

The common factors of the three terms are $2, x$ and $x$.
Therefore, $10 x^2-18 x^3+14 x^4=(2 \times x \times x \times 5)-(2 \times x \times x \times 3 \times 3 \times x)$
$
\begin{aligned}
& +(2 \times x \times x \times 7 \times x \times x) \\
= & 2 \times x \times x \times[(5-(3 \times 3 \times x)+(7 \times x \times x)] \text { (combining the three terms) }
\end{aligned}
$

Example 3:

Factorise $6 x y-4 y+6-9 x$.
Solution:
Step 1 Check if there is a common factor among all terms. There is none.
Step 2 Think of grouping. Notice that first two terms have a common factor $2 y$;
$
6 x y-4 y=2 y(3 x-2)
$

What about the last two terms? Observe them. If you change their order to $-9 x+6$, the factor $(3 x-2)$ will come out;
$
\begin{aligned}
-9 x+6 & =-3(3 x)+3(2) \\
& =-3(3 x-2)
\end{aligned}
$

Step 3 Putting (a) and (b) together,
$
\begin{aligned}
6 x y-4 y+6-9 x & =6 x y-4 y-9 x+6 \\
& =2 y(3 x-2)-3(3 x-2) \\
& =(3 x-2)(2 y-3)
\end{aligned}
$

The factors of $(6 x y-4 y+6-9 x)$ are $(3 x-2)$ and $(2 y-3)$.

Example 4:

Factorise $x^2+8 x+16$
Solution:

Observe the expression; it has three terms. Therefore, it does not fit Identity III. Also, it's first and third terms are perfect squares with a positive sign before the middle term. So, it is of the form $a^2+2 a b+b^2$ where $a=x$ and $b=4$

such that
$
\begin{aligned}
a^2+2 a b+b^2 & =x^2+2(x)(4)+4^2 \\
& =x^2+8 x+16
\end{aligned}
$

Since
$
a^2+2 a b+b^2=(a+b)^2 \text {, }
$
by comparison
$
x^2+8 x+16=(x+4)^2
$
(the required factorisation)
Example 5:

Factorise $4 y^2-12 y+9$
Solution:

Observe $4 y^2=(2 y)^2, 9=3^2$ and $12 y=2 \times 3 \times(2 y)$
Therefore,
$
4 y^2-12 y+9=(2 y)^2-2 \times 3 \times(2 y)+(3)^2
$

Observe here the given expression is of the form
$
a^2-2 a b+b^2 \text {. }
$

Where $a=2 y$, and $b=3$ with $2 a b=2 \times 2 y \times 3=12 y$.

$
=(2 y-3)^2
$
(required factorisation)
Example 6:

Factorise $49 p^2-36$
Solution:

There are two terms; both are squares and the second is negative. The expression is of the form $\left(a^2-b^2\right)$. Identity III is applicable here;
$
\begin{aligned}
49 p^2-36 & =(7 p)^2-(6)^2 \\
& =(7 p-6)(7 p+6) \text { (required factorisation) }
\end{aligned}
$

Example 7:

Factorise $a^2-2 a b+b^2-c^2$
Solution:

The first three terms of the given expression form $(a-b)^2$. The fourth term is a square. So the expression can be reduced to a difference of two squares.

$\text { Thus, } \quad a^2-2 a b+b^2-c^2=(a-b)^2-c^2 \quad \text { (Applying Identity II) }$

$\begin{array}{ll}
=[(a-b)-c)((a-b)+c)] & \text { (Applying Identity III) } \\
=(a-b-c)(a-b+c) & \text { (required factorisation) }
\end{array}$

Notice, how we applied two identities one after the other to obtain the required factorisation.
Example 8:

Factorise $m^4-256$
Solution:

We note
$
m^4=\left(m^2\right)^2 \text { and } 256=(16)^2
$

Thus, the given expression fits Identity III.
Therefore,
$
\begin{aligned}
m^4-256 & =\left(m^2\right)^2-(16)^2 \\
& =\left(m^2-16\right)\left(m^2+16\right) \quad[(\text { using Identity (III) }]
\end{aligned}
$

Now, $\left(m^2+16\right)$ cannot be factorised further, but $\left(m^2-16\right)$ is factorisable again as per Identity III.
$
\begin{aligned}
m^2-16 & =m^2-4^2 \\
& =(m-4)(m+4) \\
m^4-256 & =(m-4)(m+4)\left(m^2+16\right)
\end{aligned}
$

Example 9:

Factorise $x^2+5 x+6$
Solution:

If we compare the R.H.S. of Identity (IV) with $x^2+5 x+6$, we find $a b=6$, and $a+b=5$. From this, we must obtain $a$ and $b$. The factors then will be $(x+a)$ and $(x+b)$.

If $a b=6$, it means that $a$ and $b$ are factors of 6 . Let us try $a=6, b=1$. For these values $a+b=7$, and not 5 , So this choice is not right.
Let us try $a=2, b=3$. For this $a+b=5$ exactly as required.
The factorised form of this given expression is then $(x+2)(x+3)$.
In general, for factorising an algebraic expression of the type $x^2+p x+q$, we find two factors $a$ and $b$ of $q$ (i.e., the constant term) such that
$
a b=q \quad \text { and } \quad a+b=p
$

Then, the expression becomes or
$
\begin{array}{ll}
\text { Then, the expression becomes } & x^2+(a+b) x+a b \\
\text { or } & x^2+a x+b x+a b \\
\text { or } & x(x+a)+b(x+a)
\end{array}
$
or
$
(x+a)(x+b) \quad \text { which are the required factors. }
$

Example 10:

Find the factors of $y^2-7 y+12$.
Solution:

We note $12=3 \times 4$ and $3+4=7$. Therefore,
$
\begin{aligned}
y^2-7 y+12 & =y^2-3 y-4 y+12 \\
& =y(y-3)-4(y-3)=(y-3)(y-4)
\end{aligned}
$

Note, this time we did not compare the expression with that in Identity (IV) to identify $a$ and $b$. After sufficient practice you may not need to compare the given expressions for their factorisation with the expressions in the identities; instead you can proceed directly as we did above.

Example 11:

Obtain the factors of $z^2-4 z-12$.
Solution:

Here $a b=-12$; this means one of $a$ and $b$ is negative. Further, $a+b=-4$, this means the one with larger numerical value is negative. We try $a=-4, b=3$; but this will not work, since $a+b=-1$. Next possible values are $a=-6, b=2$, so that $a+b=-4$ as required.
Hence,
$
\begin{aligned}
z^2-4 z-12 & =z^2-6 z+2 z-12 \\
& =z(z-6)+2(z-6) \\
& =(z-6)(z+2)
\end{aligned}
$

Example 12:

Find the factors of $3 m^2+9 m+6$.
Solution:

We notice that 3 is a common factor of all the terms.
Therefore,
$
\begin{aligned}
3 m^2+9 m+6 & =3\left(m^2+3 m+2\right) \\
m^2+3 m+2 & =m^2+m+2 m+2 \\
& =m(m+1)+2(m+1) \\
& =(m+1)(m+2)
\end{aligned}
$

Now,
(as $2=1 \times 2$ )

Therefore,
$
3 m^2+9 m+6=3(m+1)(m+2)
$

Example 13:

Do the following divisions.
(i) $-20 x^4 \div 10 x^2$
(ii) $7 x^2 y^2 z^2 \div 14 x y z$

Solution:
(i) $-20 x^4=-2 \times 2 \times 5 \times x \times x \times x \times x$
$10 x^2=2 \times 5 \times x \times x$

Therefore, $\quad\left(-20 x^4\right) \div 10 x^2=\frac{-2 \times 2 \times 5 \times x \times x \times x \times x}{2 \times 5 \times x \times x}=-2 \times x \times x=-2 x^2$
(ii) $7 x^2 y^2 z^2 \div 14 x y z$
$
\begin{aligned}
& =\frac{7 \times x \times x \times y \times y \times z \times z}{2 \times 7 \times x \times y \times z} \\
& =\frac{x \times y \times z}{2}=\frac{1}{2} x y z
\end{aligned}
$

Example 14:

Divide $24\left(x^2 y z+x y^2 z+x y z^2\right)$ by $8 x y z$ using both the methods.
Solution:
$
\begin{aligned}
& 24\left(x^2 y z+x y^2 z+x y z^2\right) \\
& =2 \times 2 \times 2 \times 3 \times[(x \times x \times y \times z)+(x \times y \times y \times z)+(x \times y \times z \times z)] \\
& =2 \times 2 \times 2 \times 3 \times x \times y \times z \times(x+y+z)=8 \times 3 \times x y z \times(x+y+z)
\end{aligned}
$

Therefore, $24\left(x^2 y z+x y^2 z+x y z^2\right) \div 8 x y z$
(By taking out the
$
=\frac{8 \times 3 \times x y z \times(x+y+z)}{8 \times x y z}=3 \times(x+y+z)=3(x+y+z)
$
common factor)

$\text { Alternately, } \begin{aligned}
24\left(x^2 y z+x y^2 z+x y z^2\right) \div 8 x y z & =\frac{24 x^2 y z}{8 x y z}+\frac{24 x y^2 z}{8 x y z}+\frac{24 x y z^2}{8 x y z} \\
& =3 x+3 y+3 z=3(x+y+z)
\end{aligned}$

Example 15:

Divide $44\left(x^4-5 x^3-24 x^2\right)$ by $11 x(x-8)$
Solution:

Factorising $44\left(x^4-5 x^3-24 x^2\right)$, we get
$
44\left(x^4-5 x^3-24 x^2\right)=2 \times 2 \times 11 \times x^2\left(x^2-5 x-24\right)
$
(taking the common factor $x^2$ out of the bracket)
$
\begin{aligned}
& =2 \times 2 \times 11 \times x^2\left(x^2-8 x+3 x-24\right) \\
& =2 \times 2 \times 11 \times x^2[x(x-8)+3(x-8)] \\
& =2 \times 2 \times 11 \times x^2(x+3)(x-8)
\end{aligned}
$

Therefore, $44\left(x^4-5 x^3-24 x^2\right) \div 11 x(x-8)$
$
=\frac{2 \times 2 \times 11 \times x \times x \times(x+3) \times(x-8)}{11 \times x \times(x-8)}
$

$=2 \times 2 \times x(x+3)=4 x(x+3)$

We cancel the factors 11 , $x$ and $(x-8)$ common to both the numerator and denominator.

Example 16:

Divide $z\left(5 z^2-80\right)$ by $5 z(z+4)$

Solution:

$
\begin{aligned}
& \text {Dividend }=z\left(5 z^2-80\right) \\
& =\mathrm{z}\left[\left(5 \times z^2\right)-(5 \times 16)\right] \\
& =z \times 5 \times\left(z^2-16\right) \\
& =5 z \times(z+4)(z-4) \\
&
\end{aligned}
$
[using the identity
$
\left.a^2-b^2=(a+b)(a-b)\right]
$

Thus,
$
z\left(5 z^2-80\right) \div 5 z(z+4)=\frac{5 z(z-4)(z+4)}{5 z(z+4)}=(z-4)
$