11
BRILLIANT COACHING
UNIT-2
POLYNOMIALS
It is not once nor twice but times without number that the same ideas
make their appearance in the world.
1. Find the value for K for which x
(Ans : K= - 91)
4 + 10x3 + 25x2 + 15x + K exactly divisible by x + 7.Ans:
Since P(x) exactly divisible by g(x)
Let P(x) = x4 + 10x4 + 25x2 + 15x + K and g(x) = x + 7\
now x + 7
r (x) = 03 2
4 3 2
4 3
3 4 13
10 25 15
7
x x x
x x x x K
x x
+ + −
+ + + +
+
-------------
3x
3 + 25 x23x
3 + 21x2-------------------
4x
4x
------------------
-13x + K
- 13x - 91
----------------
K + 91
------------
2 + 15 x2 + 28x\
2. If two zeros of the polynomial f(x) = x
other zeros. (Ans:7, -5)
K + 91 = 04 - 6x3 - 26x2 + 138x – 35 are 2± Ö3.Find theAns
Sum of Zeros = 2 + 3 + 2 - 3
= 4
Product of Zeros = ( 2+ 3 )(2 - 3 )
= 4 – 3
= 1
Quadratic polynomial is x
K= -91
12
x
x
4 3 2
: Let the two zeros are 2 + 3 and 2 - 32 – (sum) x + Product2 – 2x – 352 – 4x + 1 4 3 26 26 138 35
4
x x x x
x x x
− − + −
− +
-----------------
-2x
- 2x
-----------------------
-35x
-35x
------------------------
0
------------------------
3 – 27x2 + 138x3 + 8x2 – 2x2 + 140x – 352 + 140x – 35\
(x – 7)(x + 5) = 0
x = 7, -5
3. Find the Quadratic polynomial whose sum and product of zeros are
2 1
1
x2 – 2x – 35 = 0Ö2 + 1,+
.
Ans:
Product = 1
Q.P =
X
sum = 2 22 – (sum) x + Product\
4. If
(
x2 – (2 2 ) x + 1a,b are the zeros of the polynomial 2x2 – 4x + 5 find the value of a) a2 + b 2 b)a - b)2.(Ans: a) -1 , b) –6)
Ans
: p (x) = 2 x2 – 4 x + 5a
4
2
+ b =b
a
−
=
= 2a b
5
2
=c
a
=
a
2 + b2 = (a + b)2 – 2 a bSubstitute then we get,
(
a 2+ b2 = -1a - b)2 = (a + b)2 - 4 a bSubstitute, we get = (
other two Zeros are 7 and -5
13
5. If
whose zeros are a)
a - b)2 = - 6a,b are the zeros of the polynomial x2 + 8x + 6 frame a Quadratic polynomiala
1
and
b
1
b) 1 +
a
b
, 1 +
b
a
.
(Ans: x
3
4
+
6
1
, x
3
32
+
3
32
)
2+ x2- xAns:
p (x) = x2 + 8 x + 6a
a) Let two zeros are
+ b = -8 and a b = 6a
1
and
b
1
Sum =
a
1
+
b
1
=
.
a b
a b
+
=
8
6
−
=
4
3
−
Product =
a
1
x
b
1
=
1 1
a
.b 6=
Required Q.P is
x
4 1
3 6
2 +x
+b) Let two Zeros are 1+
a
b
and 1 +
b
a
sum = 1+
a
b
+1+
b
a
= 2 +
b
a
+
a
b
= 2+
ab
a
2 + b 2= 2+
ab
(
a b ) 2ab + 2 −after solving this problem,
We get =
3
32
Product = ( 1 +
a
b
)(1+
b
a
)
= 1+
b
a
+
a
b
+1
= 2 +
ab
a
2 + b 2Substitute this sum,
14
We get =
3
32
Required Q.P. is x
3
32
x +
3
32
6. On dividing the polynomial 4x
quotient is x
(Ans:4 x
2 -4 - 5x3 - 39x2 - 46x – 2 by the polynomial g(x) the2 - 3x – 5 and the remainder is -5x + 8.Find the polynomial g(x).2+7x+2)Ans:
g(x) =
( )
( ) ( )
p(x) = g (x) q (x) + r (x)q x
p x
− r xlet p(x) = 4x
q(x) = x
now p(x) – r(x) = 4x
when
( )
( ) ( )
4 – 5x3 – 39x2 – 46x – 22 – 3x – 5 and r (x) = -5x + 84 – 5x3 – 39x2 – 41x - 10q x
p x
− r x= 4x
2 + 7x +2\
7. If the squared difference of the zeros of the quadratic polynomial
equal to 144 , find the value of p. (Ans:
g(x) = 4x2 + 7x + 2x2 + px + 45 is± 18).Ans:
Let two zeros are a and b where a > bAccording given condition
(
Let p(x) = x
a - b)2 = 1442 + px + 45a
+ b =a
−
b=
1
−
p= - p
ab
=a
c
=
1
45
= 45
now (
(
(-p)
Solving this we get p =
8. If
Quadratic polynomial having
a - b)2 = 144a + b)2 – 4 ab = 1442 – 4 (45) = 144± 18a,b are the zeros of a Quadratic polynomial such that a + b = 24, a - b = 8. Find aa and b as its zeros. (Ans: k(x2– 24x + 128))Ans:
a+b = 24a
-----------
2
15
- b = 8a = 32a
2
32
= 16,
Work the same way to
So,
Q.P is x
= x
Solve this,
it is k (x
9. If
a.
=\ a = 16a+b = 24b = 82 – (sum) x + product2 – (16+8) x + 16 x 82 – 24x + 128)a & ß are the zeroes of the polynomial 2x2 4x + 5, then find the value ofa2 + ß2 b. 1/ a + 1/ ß c. (a ß)2 d. 1/a2 + 1/ß2 e. a3 + ß3(Ans:-1,
5
4
,-6,
25
−
,-7)
4Ans:
Let p(x) = 2x2 – 4x +5a
+b =a
−
b=
2
4
= 2
ab
=a
c
=
2
5
a)
a2+b2 = (a+b)2 - 2abSubstitute to get =
b)
a2+b2 = -1a
1
+
b
1
=
ab
a
+ bsubstitute , then we get =
a
1
+
b
1
=
5
4
b) (
a-b)2 = (a+b)2 - 4 abTherefore we get, (
d)
a-b)2 = - 621
a
+
21
b
=
2 2
2ab
a
+ b=
22
5
1
−
\
2
1
a
+
2
1
b
=
25
−
e)
Substitute this,
to get,
16
10. Obtain all the zeros of the polynomial p(x) = 3x
zeroes are
4a3+b3 = (a+b)(a2+b2 - ab)a3+b3 = -74 15x3 + 17x2 +5x 6 if two 1/3 and 1/3. (Ans:3,2)11.
algorithm.
a. deg p(x) = deg q(x) b. deg q(x) = deg r(x) c. deg q(x) =
Give examples of polynomials p(x), g(x), q(x) and r(x) which satisfy the division0.12. If the ratios of the polynomial ax
3abc+a
3+3bx2+3cx+d are in AP, Prove that 2b3-2d=0Ans:
but zero are in AP
let
sum =
Let p(x) = ax3 + 3bx2 + 3cx + d and a , b , r are their three Zerosa = m – n , b = m, r = m + na+b+ r =a
−
bsubstitute this sum , to get = m=
a
−
bNow taking two zeros as sum
ab +b r +ar =a
c
(m-n)m + m(m+n) + (m + n)(m – n) =
a
3
cSolve this problem , then we get
2
2
3 3a
b
− ac= n
2Product
ab r =a
d
(m-n)m (m+n) =
a
−
d(m
2 –n2)m =a
−
d[(
a
−
b)
3 3
2 – ( )2
2
a
b
− ac] (
a
−
b) =
a
−
dSimplifying we get
2b
17
13. Find the number of zeros of the polynomial from the graph given.
(Ans:1)
14. If one zero of the polynomial 3x
zeros and the value of k (Ans k= 2/3)
3 – 3abc + a2 d = 02 - 8x +2k+1 is seven times the other, find theSelf Practice
14. If (n-k) is a factor of the polynomials
k = n +
x2+px+q & x2 + m x+n. Prove thatm p
n q
−
−
Ans :
since (n – k) is a factor of x2 + px + q\
And (n – k)
Solve this problem by yourself,
(n – k)2 + p(n- k) + q = 02 + m(n – k) + n = 0\
k = n +m p
n q
−
−
SELF PRACTICE
16. If 2, ½ are the zeros of px
17. If m, n are zeroes of ax
(Ans: a=1/2 ,c=5)
18. What must be subtracted from 8x
polynomial is exactly divisible by 4x
19. What must be added to the polynomial p(x)= x
resulting polynomial is exactly divisible by x2+5x+r, prove that p= r.2-5x+c, find the value of a and c if m + n = m.n=104 + 14x3 – 2x2 + 7x –8 so that the resulting2+3x-2. (Ans: 14x – 10)4 + 2x3 – 2x2 + x –1 so that the2+2x-3. (Ans: x-2)
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