# class 9 polynomials extra questions

Here we are trying to give the class 9 polynomials extra questions and polynomials class 9 important questions It tests the basic concepts and at the same makes the student comfortable with the questions

## Match the column

A.

B.

A.
1 -> x -1
2 -> $x^2 -x -1$
3 -> $1-3x^3$
4 -> $x^4 -3x^2+2 + 3x^3$
5 -> $x^5 - 3x^2 +1$
B.
Monomial : $3x^3$
Binomial : $x-1$
Trinomial : $x^2 -2x+1$
No apprpiate match : $x^4 -3x^2+2 + 3x^3$, $x^5 - 3x^2 +1$

## Table Type

P(x)=5x2 -3x+7

P(x)=5x2 -3x+7
P(0) = 5(0)2 -3(0)+7=7
P(1) = 5(1)2 -3(1)+7=9
P(5) = 5(5)2 -3(5)+7=117
P(-1) = 5(-1)2 -3(-1)+7=15
P(-2) = 5(-2)2 -3(-2)+7=33

## True or False statement

1. P(x) =x-1 and g(x) =x2-2x +1 . p(x) is a factor of g(x)
2. The factor of 3x2 –x-4 are (x+1)(3x-4)
3. Every linear polynomial has only one zero
4. Every real number is the zero’s of zero polynomial
5. A binomial may have degree 4
6. 0,2 are the zeroes of x2-2x
7. The degree of zero polynomial is not defined

1. True, as g(1)=0
2. True, we can get this by split method
3. True
4. True
5. True , example x4 +1
6. True
7. True

Question 1
Factorize following
1. x2 +9x+18
2. 3x3 –x2-3x+1
3. x3-23x2+142x-120
4. 1+8x3

(a) (x+6)(x+3)
(b) (3x-1)(x-1)(x+1)
(c)(x-1)(x-10)(x-12)
(d) (2x+1)(4x2-2x+1)

## Multiple choice Questions

Question 1
Find the remainder when x4+x3-2x2+x+1 is divided by x-1 (a)1
(b)5
(c)2
(d)3

( c)

Question 2
Which of these identities is not true?

(d)

Question 3
If $\frac {x}{y} + \frac {y}{x}=-1$ and ($x y, \ne 0$ ) , the value of $x^3 – y^3$ is
(a) 1
(b) -1
(c) 0
(d) 2

$\frac {x}{y} + \frac {y}{x}=-1$
$x^2 + y^2=-xy$
or
$x^2 +y^2 +xy=0$
Now
$x^3 – y^3= (x-y)(x^2 +y^2 +xy)=(x-y) \times 0=0$
Hence (c) is the correct answer

Question 4
If $p+q+r=0$ ,then the value of $\frac {p^2}{qr} + \frac {q^2}{pr} + \frac {r^2}{pq}$ is
(a) 1
(b) 3
(c) -1
(d) 0

Now we know that
$p^3 + q^3 + r^3 -3pqr=(p+q+r)(p^2 + q^2 + r^2 -pq -qr -pr)$
as $p+q+r=0$
$p^3 + q^3 + r^3 -3pqr=0$ or $p^3 + q^3 + r^3 =3pqr$
Now
$\frac {p^2}{qr} + \frac {q^2}{pr} + \frac {r^2}{pq}= \frac {p^3 + q^3 + r^3}{pqr}= \frac {3pqr}{pqr}=3$
Hence (b) is the correct answer

## class 9 polynomials important questions

Question 1
Find the value of $47^3 - 30^3 - 17^3$

$47^3 - 30^3 - 17^3=47^3 + (-30)^3 +(-17)^3$
Now we know that
$p^3 + q^3 + r^3 -3pqr=(p+q+r)(p^2 + q^2 + r^2 -pq -qr -pr)$
and if p+q+r=0,then
$p^3 + q^3 + r^3=3pqr$
Here 47+(-30)+ (-17)=0
Therefore
$47^3 - 30^3 - 17^3=47^3 + (-30)^3 +(-17)^3= 3 (47)(-30)(-17)=71910$

Question 2
Simply following
(i) $\frac {(p^2 -q^2)^3 + (q^2 -r^2)^3 + (r^2 -p^2)^3}{(p -q)^3 + (q -r)^3 + (r -p)^3}$
(ii) $(x - y)^3+ (y – z)^3+ (z - x)^3$

Now we know that
$p^3 + q^3 + r^3 -3pqr=(p+q+r)(p^2 + q^2 + r^2 -pq -qr -pr)$
and if p+q+r=0,then
$p^3 + q^3 + r^3=3pqr$
(i) Here $p^2 -q^2 + q^2 -r^2 + r^2 -p^2=0$
$p -q + q -r + r -p=0$ Therefore
$\frac {(p^2 -q^2)^3 + (q^2 -r^2)^3 + (r^2 -p^2)^3}{(p -q)^3 + (q -r)^3 + (r -p)^3}$
$=\frac {3(p^2-q^2)(q^2 -r^2)(r^2 -p^2)}{3(p-q)(q-r)(r-p)}= (p+q)(q+r)(r+p)$
(ii) Here (x - y)+ (y – z)+ (z - x)=0
Therefore
$(x - y)^3+ (y – z)^3+ (z - x)^3=3(x-y)(y-z)(z-x)$

Question 3
Find the values of a and b if $x^3 -ax^2 -13x +b=0$ has (x-1) and (x+3) as factors

Let f(x) =x^3 -ax^2 -13x +b\$
Now f(1)=0 and f(-3)=0
f(1) = 1-a -13+b=0 or b-a=12
f(-3)=9a-b-12=0
On solving both of these
a=3 and b=15

## Summary

This Polynomials class 9 extra questions is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail. You can download this test as pdf also as below