Examples (Revised) - Chapter 2 - Polynomials - Ncert Solutions class 9 - Maths

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Chapter 2 - Polynomials | NCERT Solutions for Class 9 Maths

Example 1 :

Find the degree of each of the polynomials given below:
(i) $x^5-x^4+3$
(ii) $2-y^2-y^3+2 y^8$
(iii) 2

Solution:

(i) The highest power of the variable is 5 . So, the degree of the polynomial is 5 .
(ii) The highest power of the variable is 8 . So, the degree of the polynomial is 8 .
(iii) The only term here is 2 which can be written as $2 x^0$. So the exponent of $x$ is 0 . Therefore, the degree of the polynomial is 0 .

Now observe the polynomials $p(x)=4 x+5, q(y)=2 y, r(t)=t+\sqrt{2}$ and $s(u)=3-u$. Do you see anything common among all of them? The degree of each of these polynomials is one. A polynomial of degree one is called a linear polynomial. Some more linear polynomials in one variable are $2 x-1, \sqrt{2} y+1,2-u$. Now, try and find a linear polynomial in $x$ with 3 terms? You would not be able to find it because a linear polynomial in $x$ can have at most two terms. So, any linear polynomial in $x$ will be of the form $a x+b$, where $a$ and $b$ are constants and $a \neq 0$ (why?). Similarly, $a y+b$ is a linear polynomial in $y$.
Now consider the polynomials :
$
2 x^2+5,5 x^2+3 x+\pi, x^2 \text { and } x^2+\frac{2}{5} x
$

Do you agree that they are all of degree two? A polynomial of degree two is called a quadratic polynomial. Some examples of a quadratic polynomial are $5-y^2$, $4 y+5 y^2$ and $6-y-y^2$. Can you write a quadratic polynomial in one variable with four different terms? You will find that a quadratic polynomial in one variable will have at most 3 terms. If you list a few more quadratic polynomials, you will find that any quadratic polynomial in $x$ is of the form $a x^2+b x+c$, where $a \neq 0$ and $a, b, c$ are constants. Similarly, quadratic polynomial in $y$ will be of the form $a y^2+b y+c$, provided $a \neq 0$ and $a, b, c$ are constants.

We call a polynomial of degree three a cubic polynomial. Some examples of a cubic polynomial in $x$ are $4 x^3, 2 x^3+1,5 x^3+x^2, 6 x^3-x, 6-x^3, 2 x^3+4 x^2+6 x+7$. How many terms do you think a cubic polynomial in one variable can have? It can have at most 4 terms. These may be written in the form $a x^3+b x^2+c x+d$, where $a \neq 0$ and $a, b, c$ and $d$ are constants.

Now, that you have seen what a polynomial of degree 1, degree 2, or degree 3 looks like, can you write down a polynomial in one variable of degree $n$ for any natural number $n$ ? A polynomial in one variable $x$ of degree $n$ is an expression of the form
$
a_n x^n+a_{n-1} x^{n-1}+\ldots+a_1 x+a_0
$

where $a_0, a_1, a_2, \ldots, a_n$ are constants and $a_n \neq 0$.
In particular, if $a_0=a_1=a_2=a_3=\ldots=a_n=0$ (all the constants are zero), we get the zero polynomial, which is denoted by 0 . What is the degree of the zero polynomial? The degree of the zero polynomial is not defined.

So far we have dealt with polynomials in one variable only. We can also have polynomials in more than one variable. For example, $x^2+y^2+x y z$ (where variables are $x, y$ and $z$ ) is a polynomial in three variables. Similarly $p^2+q^{10}+r$ (where the variables are $p, q$ and $r$ ), $u^3+v^2$ (where the variables are $u$ and $v$ ) are polynomials in three and two variables, respectively. You will be studying such polynomials in detail later.

Example 2;

Find the value of each of the following polynomials at the indicated value of variables:
(i) $p(x)=5 x^2-3 x+7$ at $x=1$.
(ii) $q(y)=3 y^3-4 y+\sqrt{11}$ at $y=2$.
(iii) $p(t)=4 t^4+5 t^3-t^2+6$ at $t=a$.

Solution :

(i) $p(x)=5 x^2-3 x+7$
The value of the polynomial $p(x)$ at $x=1$ is given by
$
\begin{aligned}
p(1) & =5(1)^2-3(1)+7 \\
& =5-3+7=9
\end{aligned}
$
(ii) $q(y)=3 y^3-4 y+\sqrt{11}$

The value of the polynomial $q(y)$ at $y=2$ is given by
$
q(2)=3(2)^3-4(2)+\sqrt{11}=24-8+\sqrt{11}=16+\sqrt{11}
$
(iii) $p(t)=4 t^4+5 t^3-t^2+6$

The value of the polynomial $p(t)$ at $t=a$ is given by
$
p(a)=4 a^4+5 a^3-a^2+6
$

Now, consider the polynomial $p(x)=x-1$.
What is $p(1)$ ? Note that : $p(1)=1-1=0$.
As $p(1)=0$, we say that 1 is a zero of the polynomial $p(x)$.
Similarly, you can check that 2 is a zero of $q(x)$, where $q(x)=x-2$.
In general, we say that a zero of a polynomial $p(x)$ is a number $c$ such that $p(c)=0$.
You must have observed that the zero of the polynomial $x-1$ is obtained by equating it to 0 , i.e., $x-1=0$, which gives $x=1$. We say $p(x)=0$ is a polynomial equation and 1 is the root of the polynomial equation $p(x)=0$. So we say 1 is the zero of the polynomial $x-1$, or a root of the polynomial equation $x-1=0$.

Now, consider the constant polynomial 5. Can you tell what its zero is? It has no zero because replacing $x$ by any number in $5 x^0$ still gives us 5 . In fact, a non-zero constant polynomial has no zero. What about the zeroes of the zero polynomial? By convention, every real number is a zero of the zero polynomial.

Example 3 :

Check whether -2 and 2 are zeroes of the polynomial $x+2$.
Solution :

Let $p(x)=x+2$.
Then $p(2)=2+2=4, p(-2)=-2+2=0$
Therefore, -2 is a zero of the polynomial $x+2$, but 2 is not.
Example 4 :

Find a zero of the polynomial $p(x)=2 x+1$.
Solution :

Finding a zero of $p(x)$, is the same as solving the equation
$
p(x)=0
$

Now,
$2 x+1=0$ gives us $x=-\frac{1}{2}$

So, $-\frac{1}{2}$ is a zero of the polynomial $2 x+1$.
Now, if $p(x)=a x+b, a \neq 0$, is a linear polynomial, how can we find a zero of $p(x)$ ? Example 4 may have given you some idea. Finding a zero of the polynomial $p(x)$, amounts to solving the polynomial equation $p(x)=0$.
Now, $p(x)=0$ means
$
\begin{aligned}
a x+b & =0, a \neq 0 \\
a x & =-b \\
x & =-\frac{b}{a} .
\end{aligned}
$

So,
i.e.,

So, $x=-\frac{b}{a}$ is the only zero of $p(x)$, i.e., a linear polynomial has one and only one zero.
Now we can say that 1 is the zero of $x-1$, and -2 is the zero of $x+2$.
Chapter 2 - Polynomials | NCERT Solutions for Class 9 Maths

Verify whether 2 and 0 are zeroes of the polynomial $x^2-2 x$.
Solution :

Let
$
\begin{aligned}
& p(x)=x^2-2 x \\
& p(2)=2^2-4=4-4=0 \\
& p(0)=0-0=0
\end{aligned}
$

Then
and
Hence, 2 and 0 are both zeroes of the polynomial $x^2-2 x$.
Let us now list our observations:
(i) A zero of a polynomial need not be 0 .
(ii) 0 may be a zero of a polynomial.
(iii) Every linear polynomial has one and only one zero.
(iv) A polynomial can have more than one zero.

Example 6 :

Examine whether $x+2$ is a factor of $x^3+3 x^2+5 x+6$ and of $2 x+4$.
Solution:

The zero of $x+2$ is -2 . Let $p(x)=x^3+3 x^2+5 x+6$ and $s(x)=2 x+4$
Then,
$
\begin{aligned}
p(-2) & =(-2)^3+3(-2)^2+5(-2)+6 \\
& =-8+12-10+6 \\
& =0
\end{aligned}
$

So, by the Factor Theorem, $x+2$ is a factor of $x^3+3 x^2+5 x+6$.
Again,
$
s(-2)=2(-2)+4=0
$

So, $x+2$ is a factor of $2 x+4$. In fact, you can check this without applying the Factor Theorem, since $2 x+4=2(x+2)$.

Example 7 :

Find the value of $k$, if $x-1$ is a factor of $4 x^3+3 x^2-4 x+k$.
Solution :

As $x-1$ is a factor of $p(x)=4 x^3+3 x^2-4 x+k, p(1)=0$
Now,
$
p(1)=4(1)^3+3(1)^2-4(1)+k
$

So,
$
\begin{aligned}
4+3-4+k & =0 \\
k & =-3
\end{aligned}
$
i.e.,

We will now use the Factor Theorem to factorise some polynomials of degree 2 and 3 . You are already familiar with the factorisation of a quadratic polynomial like $x^2+l x+m$. You had factorised it by splitting the middle term $l x$ as $a x+b x$ so that $a b=m$. Then $x^2+l x+m=(x+a)(x+b)$. We shall now try to factorise quadratic polynomials of the type $a x^2+b x+c$, where $a \neq 0$ and $a, b, c$ are constants.
Factorisation of the polynomial $a x^2+b x+c$ by splitting the middle term is as follows:
Let its factors be $(p x+q)$ and $(r x+s)$. Then $\quad \frac{3 x^2}{x}=3 x=$ first term of quotient
$
a x^2+b x+c=(p x+q)(r x+s)=p r x^2+(p s+q r) x+q s
$

Comparing the coefficients of $x^2$, we get $a=p r$.
Similarly, comparing the coefficients of $x$, we get $b=p s+q r$.
And, on comparing the constant terms, we get $c=q s$.
This shows us that $b$ is the sum of two numbers $p s$ and $q r$, whose product is $(p s)(q r)=(p r)(q s)=a c$

Therefore, to factorise $a x^2+b x+c$, we have to write $b$ as the sum of two numbers whose product is ac. This will be clear from Example 13.

Example 8 :

Factorise $6 x^2+17 x+5$ by splitting the middle term, and by using the Factor Theorem.

Solution 1 :

(By splitting method) : If we can find two numbers $p$ and $q$ such that $p+q=17$ and $p q=6 \times 5=30$, then we can get the factors.
So, let us look for the pairs of factors of 30 . Some are 1 and 30,2 and 15,3 and 10,5 and 6 . Of these pairs, 2 and 15 will give us $p+q=17$.

So, $6 x^2+17 x+5=6 x^2+(2+15) x+5$
$
\begin{aligned}
& =6 x^2+2 x+15 x+5 \\
& =2 x(3 x+1)+5(3 x+1) \\
& =(3 x+1)(2 x+5)
\end{aligned}
$

Solution 2 :

(Using the Factor Theorem)
$6 x^2+17 x+5=6\left(x^2+\frac{17}{6} x+\frac{5}{6}\right)=6 p(x)$, say. If $a$ and $b$ are the zeroes of $p(x)$, then $6 x^2+17 x+5=6(x-a)(x-b)$. So, $a b=\frac{5}{6}$. Let us look at some possibilities for $a$ and
b. They could be $\pm \frac{1}{2}, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm \frac{5}{2}, \pm 1$. Now, $p\left(\frac{1}{2}\right)=\frac{1}{4}+\frac{17}{6}\left(\frac{1}{2}\right)+\frac{5}{6} \neq 0$. But $p\left(\frac{-1}{3}\right)=0$. So, $\left(x+\frac{1}{3}\right)$ is a factor of $p(x)$. Similarly, by trial, you can find that $\left(x+\frac{5}{2}\right)$ is a factor of $p(x)$.

Therefore,
$
\begin{aligned}
6 x^2+17 x+5 & =6\left(x+\frac{1}{3}\right)\left(x+\frac{5}{2}\right) \\
& =6\left(\frac{3 x+1}{3}\right)\left(\frac{2 x+5}{2}\right) \\
& =(3 x+1)(2 x+5)
\end{aligned}
$

For the example above, the use of the splitting method appears more efficient. However, let us consider another example.

Example 9 :

Factorise $y^2-5 y+6$ by using the Factor Theorem.
Solution :

Let $p(y)=y^2-5 y+6$. Now, if $p(y)=(y-a)(y-b)$, you know that the constant term will be $a b$. So, $a b=6$. So, to look for the factors of $p(y)$, we look at the factors of 6 .
The factors of 6 are 1,2 and 3 .
Now, $p(2)=2^2-(5 \times 2)+6=0$
So, $y-2$ is a factor of $p(y)$.

Also, $p(3)=3^2-(5 \times 3)+6=0$
So, $y-3$ is also a factor of $y^2-5 y+6$.
Therefore, $y^2-5 y+6=(y-2)(y-3)$
Note that $y^2-5 y+6$ can also be factorised by splitting the middle term $-5 y$.
Now, let us consider factorising cubic polynomials. Here, the splitting method will not be appropriate to start with. We need to find at least one factor first, as you will see in the following example.

Example 10 :

Factorise $x^3-23 x^2+142 x-120$.
Solution :

Let $p(x)=x^3-23 x^2+142 x-120$
We shall now look for all the factors of -120 . Some of these are $\pm 1, \pm 2, \pm 3$,
$
\pm 4, \pm 5, \pm 6, \pm 8, \pm 10, \pm 12, \pm 15, \pm 20, \pm 24, \pm 30, \pm 60 \text {. }
$

By trial, we find that $p(1)=0$. So $x-1$ is a factor of $p(x)$.
Now we see that $x^3-23 x^2+142 x-120=x^3-x^2-22 x^2+22 x+120 x-120$
$
\begin{aligned}
& =x^2(x-1)-22 x(x-1)+120(x-1) \quad \text { (Why?) } \\
& =(x-1)\left(x^2-22 x+120\right) \quad \text { [Taking }(x-1) \text { common] }
\end{aligned}
$

We could have also got this by dividing $p(x)$ by $x-1$.
Now $x^2-22 x+120$ can be factorised either by splitting the middle term or by using the Factor theorem. By splitting the middle term, we have:
$
\begin{aligned}
x^2-22 x+120 & =x^2-12 x-10 x+120 \\
& =x(x-12)-10(x-12) \\
& =(x-12)(x-10)
\end{aligned}
$

So,
$
x^3-23 x^2-142 x-120=(x-1)(x-10)(x-12)
$

Example 11 :

Find the following products using appropriate identities:
(i) $(x+3)(x+3)$
(ii) $(x-3)(x+5)$

Solution :

(i) Here we can use Identity I : $(x+y)^2=x^2+2 x y+y^2$. Putting $y=3$ in it, we get
$
\begin{aligned}
(x+3)(x+3) & =(x+3)^2=x^2+2(x)(3)+(3)^2 \\
& =x^2+6 x+9
\end{aligned}
$
(ii) Using Identity IV above, i.e., $(x+a)(x+b)=x^2+(a+b) x+a b$, we have
$
\begin{aligned}
(x-3)(x+5) & =x^2+(-3+5) x+(-3)(5) \\
& =x^2+2 x-15
\end{aligned}
$

Example 12 :

Evaluate $105 \times 106$ without multiplying directly.
Solution :
$
\begin{aligned}
105 \times 106 & =(100+5) \times(100+6) \\
& =(100)^2+(5+6)(100)+(5 \times 6), \text { using Identity IV } \\
& =10000+1100+30 \\
& =11130
\end{aligned}
$

You have seen some uses of the identities listed above in finding the product of some given expressions. These identities are useful in factorisation of algebraic expressions also, as you can see in the following examples.
Example 13 :

Factorise:
(i) $49 a^2+70 a b+25 b^2$
(ii) $\frac{25}{4} x^2-\frac{y^2}{9}$

Solution :

(i) Here you can see that
$
49 a^2=(7 a)^2, 25 b^2=(5 b)^2, 70 a b=2(7 a)(5 b)
$

Comparing the given expression with $x^2+2 x y+y^2$, we observe that $x=7 a$ and $y=5 b$. Using Identity I, we get
$
49 a^2+70 a b+25 b^2=(7 a+5 b)^2=(7 a+5 b)(7 a+5 b)
$
(ii) We have $\frac{25}{4} x^2-\frac{y^2}{9}=\left(\frac{5}{2} x\right)^2-\left(\frac{y}{3}\right)^2$

Now comparing it with Identity III, we get
$
\begin{aligned}
\frac{25}{4} x^2-\frac{y^2}{9} & =\left(\frac{5}{2} x\right)^2-\left(\frac{y}{3}\right)^2 \\
& =\left(\frac{5}{2} x+\frac{y}{3}\right)\left(\frac{5}{2} x-\frac{y}{3}\right)
\end{aligned}
$

So far, all our identities involved products of binomials. Let us now extend the Identity I to a trinomial $x+y+z$. We shall compute $(x+y+z)^2$ by using Identity I.
Let $x+y=t$. Then,
$
(x+y+z)^2=(t+z)^2
$

$
=t^2+2 t z+t^2
$
(Using Identity I)

$
=(x+y)^2+2(x+y) z+z^2
$
(Substituting the value of $t$ )

$=x^2+2 x y+y^2+2 x z+2 y z+z^2$
(Using Identity I)

$=x^2+y^2+z^2+2 x y+2 y z+2 z x \quad \text { (Rearranging the terms) }$

So, we get the following identity:
Identity V : $(x+y+z)^2=x^2+y^2+z^2+2 x y+2 y z+2 z x$
Remark : We call the right hand side expression the expanded form of the left hand side expression. Note that the expansion of $(x+y+z)^2$ consists of three square terms and three product terms.

Example 14 :

Write $(3 a+4 b+5 c)^2$ in expanded form.
Solution :

Comparing the given expression with $(x+y+z)^2$, we find that $x=3 a, y=4 b$ and $z=5 c$.

Therefore, using Identity $\mathrm{V}$, we have
$
\begin{aligned}
(3 a+4 b+5 c)^2 & =(3 a)^2+(4 b)^2+(5 c)^2+2(3 a)(4 b)+2(4 b)(5 c)+2(5 c)(3 a) \\
& =9 a^2+16 b^2+25 c^2+24 a b+40 b c+30 a c
\end{aligned}
$

Example 15 :

Expand $(4 a-2 b-3 c)^2$.
Solution :

Using Identity V, we have
$
\begin{aligned}
(4 a-2 b-3 c)^2 & =[4 a+(-2 b)+(-3 c)]^2 \\
& =(4 a)^2+(-2 b)^2+(-3 c)^2+2(4 a)(-2 b)+2(-2 b)(-3 c)+2(-3 c)(4 a) \\
& =16 a^2+4 b^2+9 c^2-16 a b+12 b c-24 a c
\end{aligned}
$

Example 16 :

Factorise $4 x^2+y^2+z^2-4 x y-2 y z+4 x z$.
Solution :

We have $4 x^2+y^2+z^2-4 x y-2 y z+4 x z=(2 x)^2+(-y)^2+(z)^2+2(2 x)(-y)$ $+2(-y)(z)+2(2 x)(z)$
$
\begin{aligned}
& =[2 x+(-y)+z]^2 \quad \text { (Using Identity V) } \\
& =(2 x-y+z)^2=(2 x-y+z)(2 x-y+z)
\end{aligned}
$

So far, we have dealt with identities involving second degree terms. Now let us extend Identity I to compute $(x+y)^3$. We have:
$
(x+y)^3=(x+y)(x+y)^2
$

$\begin{aligned}
& =(x+y)\left(x^2+2 x y+y^2\right) \\
& =x\left(x^2+2 x y+y^2\right)+y\left(x^2+2 x y+y^2\right) \\
& =x^3+2 x^2 y+x y^2+x^2 y+2 x y^2+y^3 \\
& =x^3+3 x^2 y+3 x y^2+y^3 \\
& =x^3+y^3+3 x y(x+y)
\end{aligned}$

So, we get the following identity:
Identity VI : $\quad(x+y)^3=x^3+y^3+3 x y(x+y)$
Also, by replacing $y$ by $-y$ in the Identity VI, we get
Identity VII : $(x-y)^3=x^3-y^3-3 x y(x-y)$
$
=x^3-3 x^2 y+3 x y^2-y^3
$

Example 17 :

Write the following cubes in the expanded form:
(i) $(3 a+4 b)^3$
(ii) $(5 p-3 q)^3$

Solution :

(i) Comparing the given expression with $(x+y)^3$, we find that
$
x=3 a \text { and } y=4 b .
$

So, using Identity VI, we have:
$
\begin{aligned}
(3 a+4 b)^3 & =(3 a)^3+(4 b)^3+3(3 a)(4 b)(3 a+4 b) \\
& =27 a^3+64 b^3+108 a^2 b+144 a b^2
\end{aligned}
$
(ii) Comparing the given expression with $(x-y)^3$, we find that
$
x=5 p, y=3 q \text {. }
$

So, using Identity VII, we have:
$
\begin{aligned}
(5 p-3 q)^3 & =(5 p)^3-(3 q)^3-3(5 p)(3 q)(5 p-3 q) \\
& =125 p^3-27 q^3-225 p^2 q+135 p q^2
\end{aligned}
$

Example 18 :

Evaluate each of the following using suitable identities:
(i) $(104)^3$
(ii) $(999)^3$

Solution :

(i) We have
$
\begin{aligned}
(104)^3 & =(100+4)^3 \\
& =(100)^3+(4)^3+3(100)(4)(100+4)
\end{aligned}
$
(Using Identity VI)

Example 19 :

Factorise $8 x^3+27 y^3+36 x^2 y+54 x y^2$
Solution :

The given expression can be written as
$
\begin{aligned}
(2 x)^3+(3 y)^3 & +3\left(4 x^2\right)(3 y)+3(2 x)\left(9 y^2\right) \\
& =(2 x)^3+(3 y)^3+3(2 x)^2(3 y)+3(2 x)(3 y)^2 \\
& =(2 x+3 y)^3 \quad \text { (Using Identity VI) } \\
& =(2 x+3 y)(2 x+3 y)(2 x+3 y)
\end{aligned}
$
(Using Identity VI)

Now consider $(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
On expanding, we get the product as
$
\begin{gathered}
x\left(x^2+y^2+z^2-x y-y z-z x\right)+y\left(x^2+y^2+z^2-x y-y z-z x\right) \\
+z\left(x^2+y^2+z^2-x y-y z-z x\right)=x^3+x y^2+x z^2-x^2 y-x y z-z x^2+x^2 y \\
+y^3+y z^2-x y^2-y^2 z-x y z+x^2 z+y^2 z+z^3-x y z-y z^2-x z^2 \\
=x^3+y^3+z^3-3 x y z \quad \text { (On simplification) }
\end{gathered}
$

So, we obtain the following identity:
Identity VIII : $x^3+y^3+z^3-3 x y z=(x+y+z)\left(x^2+y^2+z^2-x y-y z-z x\right)$
Example 20 :

Factorise : $8 x^3+y^3+27 z^3-18 x y z$
Solution :

Here, we have
$
\begin{aligned}
& 8 x^3+y^3+27 z^3-18 x y z \\
&=(2 x)^3+y^3+(3 z)^3-3(2 x)(y)(3 z) \\
&=(2 x+y+3 z)\left[(2 x)^2+y^2+(3 z)^2-(2 x)(y)-(y)(3 z)-(2 x)(3 z)\right] \\
&=(2 x+y+3 z)\left(4 x^2+y^2+9 z^2-2 x y-3 y z-6 x z\right)
\end{aligned}
$