Graph of polynomial 
Number of Zeros 

0 

1 

2 

3 

4 

5 

6 

7 
a. it cuts the xaxis at two points ,so 2 zeroes
b. it cuts the xaxis at four points ,so 4 zeroes
c. Since it does not cut the axis, so 0 zeroes
d. it cuts the xaxis at 1 points ,so 1 zero’s
e. it cuts the xaxis at 1 points ,so 1 zero’s
f. Since it does not cut the axis, so 0 zeroes
g. Since it does not cut the axis, so 0 zeroes
h. it cuts the xaxis at two points ,so 2 zeroes
Graph of polynomial 
Type of polynomial 

Linear polynomial 

Quadratic polynomial 

Cubic polynomial 

Constant polynomial 







a. Quadratic as parabola
b. Three zeroes,So cubic polynomial
c. Constant value polynomial
d. Linear polynomial
e. One zeroes but not straight line. So no appropriate match found
f. Quadratic as parabola
g. Quadratic as parabola
e. Cubic as has three zeroes ,two of them same
Answer is ( c)
Let P(x) =x^{4}+x^{3}2x^{2}+x+1
Remainder when divide by x1
P(1) = 1+ 12+1+1=2
Answer is (d)
a^{3} + b^{3}= (a+b) (a^{2}+b^{2}ab)=(a+b) {(a+b)^{2} 3ab}
Now a+b=(11)/1=11
ab=30
So a^{3}+b^{3}=11( 121 90)=341
S(2)=4p+0+2=0 => p=1/2
P(x) =(x11)(x2)
Dividing p(x) by q(x) ,we get the remainder
(2k9)x (8k)k +10
Comparing this with (x+a)
We get
K=5 and a=5
(a)
Given that, the zeroes of the quadratic polynomial ax^{2} + bx + c , c ≠ 0 are equal
Value of discriminant (D) has to be zero
b^{2}– 4ac = 0
b^{2} = 4ac
Since L.H.S. (b^{2}) can never be negative
R.H.S. also can never be negative.
a and c must have same sign
$p(x)=6x^2  7x 3$
$p(x) = 6x^2 9x + 2x 3 = 3x(2x 3) + 1(2x3)=(3x+1)(2x3)
So roots are (3/2) , (1/3)
Hence option(a) is correct
">
$p(x)=x^2 + 7x +10$
$p(x) = x^2 + 5x +2x +10 = x(x +5) + 2(x+5)=(x+2)(x+5)
So roots are (2) , (5)
Hence option(d) is correct
a.(x+6)(x+3)
b.(3x1)(x1)(x+1)
c.(x1)(x10)(x12)
d. (2x+1)(4x^{2}2x+1)
a. r(x)=x^{3}x^{2} +4x5 w(x)=14
b. r(x)=x^{2}x w(x) =x+4
If $\alpha$ and $\beta$ are the zeroes of the polynomials
$x^2  (\alpha + \beta) + \alpha \beta$
Here $\alpha + \beta =5 + \sqrt {2} + 5  \sqrt {2} =10$
$\alpha \beta =(5 + \sqrt {2})(5  \sqrt {2}) = 25 4 =21$
Therefore quadratic Polynomial is
$x^2  10x + 21$
Here $a + b= \frac {4}{k}$
$ab=\frac {4}{k}$
Now $(a+b )^2 2ab= 24$
$(\frac {4}{k})^2 2 \frac {4}{k}=24$
$168k=24k^2$ or $3K^2 +k2=0$
$k=1$ or $k=\frac {2}{3}$
Let a be one of zero, then 1/a be the another zero
Now $a \times \frac {1}{a} =\frac {p}{3}$
or p=3
Here $ p + q=\frac {3}{k}$
$pq=\frac {2k}{k}$
Now
$p+q=pq$
$\frac {3}{k} = 2$
$k= \frac {3}{2}$
Here $a + b=  \frac {6}{1} = 6$ (1)
$ab= k$ (3)
Given $3a+2b= 20$ (2)
Solving (1) and (2)
a=14 and b=8
Now substituting these values in (3)
k=112
Since 2 is zero of the quadratic polynomial
$2^2 + 3 \times 2 + k=0$
k=10
Now $x^2 + 8x + 16$
$= x^2 + 4x + 4x + 16 = x(x+4) + 4(x+ 4) = (x+4)^2$
So zeroes are 4 ,4
Now if $k_1$ ,$k_2$ are the roots of quadratic polynomials $ax^2 + bx + c$, then relationship between them is given by
$k_1 + k_2 = \frac {b}{a}$ and $k_1 k_2= \frac {c}{a}$
Now lets verify the relationship in the above polynomial
$4 + (4) = \frac {8}{1}$
$8=8$
Also
$(4) \times (4) = \frac {16}{1}$
16=16
Here
$a + b= a$ or $2a + b=0$ (1)
Also $ab= b$ or $a=1$
Substituting the value of a in (1)
b=2
Hence a=1 and b=2
Here $m + n= = \frac {11}{3}$
$mn= \frac {4}{3}$
Now
$ \frac {m}{n} + \frac {n}{m} = \frac {m^2 + n^2}{mn} = \frac {(m+n)^2 2mn}{mn} = \frac {145}{12}$
$4 \sqrt {3} x^2 + 5x  2 \sqrt {3}$
$= 4 \sqrt {3} x^2 + 8x 3x  2 \sqrt {3}$
$= 4x (\sqrt {3}x + 2)  \sqrt {3}(\sqrt {3}x + 2)$
$= (4x  \sqrt {3})(\sqrt {3}x + 2)$
Here $ p + q = 5$ (1)
$pq= k$ (2)
Given $pq= 1$  (3)
From (1) and (3)
p=3, q=2
Substituting these values in (2)
k=6
Here $ a + b=\frac {4}{1} =4$
$ab=3$
$a^4 b^3 + a^3 b^4 = a^3b^3(a +b) = 27 \times 4 = 108$
Here
$a + b =p$ and $ab=q$
Now LHS
$= \frac {a^2}{b^2} + \frac {b^2}{a^2}$
$= (\frac {a}{b} + \frac {b}{a})^2 2$
$= ( \frac {a^2 + b^2}{ab})^2 2 $
$=(\frac {(a+b)^2  2ab}{ab})^2 2 = (\frac {p^2 2q}{q})^2 2$
$=\frac {p^4 +4q^2 4p^2q}{q^2} 2 = \frac {p^4}{q^2} +4  4\frac {p^2}{q} 2$
$= \frac {p^4}{q^2}  4\frac {p^2}{q} +2$
=RHS
If √3 and √3 are zeroes of the polynomial, then $(x \sqrt {3})(x+ \sqrt {3}) = x^2 3$ should be factor of it.
$x^3  4x^2  3x + 12= x^2(x4) 3(x 4) = (x^2 3)(x4)$
Clearly the other zero is 4
Here $l+m= \frac {5}{2}$
$lm= \frac {7}{2}$
Now the polynomial whose zeroes are $2l+ 3$ and $2m+ 3$ is given by
$x^2  (2l+3 + 2m + 3)x +(2l+3)(2m+3)=x^2 2(l+m)x 6x + 4lm + 6l + 6m +9$
$=x^2 10x 6x + 14 + 6(l+m) +9$
$=x^2 16x +14+15+9=x^2 16x + 38$
Here $pq + qr+ pr=\frac {5}{6}= \frac {5}{6}$
$pqr= \frac {1}{6}$
Now
$\frac {1}{p} + \frac {1}{q} + \frac {1}{r}$
$= \frac {pq+ qr + pr}{pqr} =5$
Given Roots of given polynomial is in AP
let pq, p, and p+q are the roots of
$f( x) = ax^3 + 3bx^2 + 3cx + d$
Now we know that
$sum \; of \; roots =  \frac {coefficient \; of \; x^2} {coefficient \;of\; x^3}$
$pq +p + p+q =  \frac {3b}{a}$
or $3p =  \frac {3b}{a}$
or $p =  \frac {b}{a}$ (1)
Also,
$sum \; of \; products \; of \; two roots = \frac {coefficient \; of\; x}{coefficient \; of \; x^3}$
$ (p q)\times p + p \times (p+q) + (pq)(p+q)=\frac {3c}{a}$
$3p^2  q^2 = \frac {3c}{a}$ (2)
Also,
$products \; of \; all \; roots = \frac {constant }{coefficient \; of \; x^3}$
$(pq)\times p \times (p+q) = \frac {d}{a}$
$p^3  pq^2 = \frac {d}{a}$ (3)
Substituting the value of p from (1) in (2), we get the value of q as
$ 3 (\frac {b}{a})^2  q^2 = \frac {3c}{a}$
or
$q^2 = 3 \frac {b^2}{a^2}  3 \frac {c}{a}$ (4)
Now that we have got the values of p and q, substituting these in equation (3)
$( \frac {b}{a})^3 +\frac {b}{a}( 3 \frac {b^2}{a^2} 3\frac {c}{a}) = \frac {d}{a}$
$ \frac {b^3}{a^3} + \frac {b}{a}(3 \frac {b^2}{a^2} 3\frac {c}{a} ) = \frac {d}{a}$
$ \frac {b^3}{a^3} +3 \frac {b^3}{a^3} 3 \frac {bc}{a^2} = \frac {d}{a}$
$2\frac {b^3}{a^3} 3 \frac {bc}{a^2} = \frac {d}{a}$
$2b^3 3abc + a^2d = 0$
If √(3/2) and √(3/2) are zeroes of the polynomial, then $(x \sqrt {3/2})(x+ \sqrt {3/2}) = 2x^2 3$ should be factor of it.
$f(x) = 2x^4  2x^3  7x^2 + 3x + 6$
Lets use division algorithm to find the other factors
So,
$2x^4  2x^3  7x^2 + 3x + 6= (2x^2 3)(x^2 x 2) = (2x^2 3)(x^2 2x + x2) =(2x^2 3)(x2)(x+1)$
So other zeros are x=2 and x=1
i. 1
ii. 2
iii. 3
iv 1
v. 4
This Class 10 Maths Extra Questions for Polynomials with answers 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.