EXERCISE 2.1
1. The
graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each
case.
Answer
(i)the graph does not intersect
the x axis so that there are no zero in this graph
(ii)the graph intersects
the x axis at one places so that here are one zeros in this graph
(iii)the graph intersects the x axis at three places so that
there are three zeros in this graph
(iv)the graph intersects
the x axis at two places so there are two zeros in this graph
(v) the graph intersect
the x axis at four places so there are four zeros in this graph
(vi) the graph intersects the x axis at three places so that
there are three zeros in this graph
EXERCISE 2.2
1. Find the
zeroes of the following quadratic polynomials and verify the relationship
between the zeroes and the coefficients.
(i) x2 – 2x – 8 (ii) 4s2 – 4s + 1 (iii) 6x2 – 3 – 7x
(iv) 4u2 + 8u (v) t2 – 15 (vi) 3x2
– x – 4
(i)
Factorize
the equation
Compare the equation
with ax^2 + bx + c =
0
We get
a = 1 ,b=-2
c= -8
To
factorize the value we have to find two value which
sum is
equal to b =-2
and product is
a*c = 1*-8 = -8
2 and -4
are such number which
sum is 2-4 = - 2
product is 2*-4 = - 8
So we can
write middle term -2 x = 2 x – 4 x
We get
x (x +2) -4(x+2) = 0
(x+2)(x-4) = 0
First zero
X+2 = 0
X = - 2
Second zero
x-4 = 0
x = 4
sum of zero -2
+ 4 = 2
product of zero
2*-4 = -8
For a equation ax^2 + bx + c =
0 , if zeroare α and β ,
Plug the
values of a , b and c we get
Sum of zeros -b/a = -(-2/1) = 2
Product of zeros c/a =-8/1
= -8
(ii) 4s2 – 4s + 1
Factorize the equation
Compare the
equation with as^2 + bs + c =
0
We get
a = 4 ,b=-4
c=1
To
factorize the value we have to find two value which
sum is
equal to b =-4
product
is a*c = 4
-2 and -2
are such number which
sum is –
2 – 2 =
- 4
product is - 2 * - 2 =
4
So we can
write middle term - 4 s = - 2 s – 2 s
We get
2s (2s -1) -1(2s-1) = 0
(2s-1)(2s-1) =
0
Solve for
first zero
2s-1 =0
2s = 1
s = ½
Solve for
second zero
2s-1 =0
2s = 1
S = ½
sum of zeros
1 /2 + 1/2 =
2/2 =
1
product of zeros ½* ½ =
1/4
Sum of zero
= -b/a = -(-4/4) = 1
Product of zero=
c/a = 1/4 = 1/4
(iii) 6x2 – 3 – 7x
Factorize the equation
Compare the
equation with ax^2 + bx + c =
0
We get
a = 6 ,b=-7
c= -3
To
factorize the value we have to find two value which
sum is
equal to b =-7
product
is a*c = 6*-3 =
-18
2 and -9
are such number which
sum is 2 – 9 = - 7
product is 2 * - 9 =
- 18
So we can
write middle term -7x =
2x – 9x
We get
2x (3x +1)
-3(3x +1) = 0
(3x+1)(2x-3) = 0
3x+1 = 0 ,
2x-3 = 0
Solve for
first zero
3x = - 1
x = -1/3
Solve for
second zero
2x-3 = 0
2x = 3
X = 3/2
Plug the
values of a , b and c we get
Sum of zeros
-b/a = -(-7/6) = 7/6
Product of zero c/a = -3/6 =
-1/2
(iv) 4u2
+ 8u
Factorize
the equation
Compare the
equation with au^2 + bu + c = 0
We get
a = 4 ,b=-8
c= 0
To
factorize the value we can take 4u
common there
4u(u+2) = 0
First zero
4u = 0
U = 0
second zero
u+2 = 0
u = - 2
sum of zero 0 - 2 = - 2
product of zero 0 * ( - 2 ) = 0
Sum of zero -b/a = -(8/4) = -2
Product of zero c/a = 0/4 =
0
(v) t2 – 15
Factorize the equation
Add 15 both side we get
Take square
root both side
First zero is
√15
second zerois -
√15
Sum of zero
√15 - √15 =
0
Product of zero √15*-√15 = -15
Compare the
equation with at^2 + bt +c = 0
We get
a = 1 ,b=0
c= -15
Sum of zero -b/a
= -(0/1) = 0
Product of zero c/a = -15/1 =
-15
(vi) 3x2 – x – 4
Factorize
the equation
Compare the
equation with ax^2 + bx +
c = 0
We get
a = 3 ,b=-1
c= -4
to
factorize the value we have to find two value which
sum is
equal to b =-1
product
is a*c =
3*-4 = -12
3 and -4
are such number which
Sum is 3 – 4 = - 1
Product is 3 *- 4 =
- 12
So we can
write middle term -x = 3x – 4x
We get
3x (x +1) -4(x +1) = 0
( x + 1)( 3x - 4) = 0
x+1 = 0 , 3x-4 = 0
Solve for
first zero
x =
- 1
solve for
second zero
3x-4 = 0
3x = 4
x = 4/3
Sum of zeros
-b/a =
-(-1/3) = 1/3
Product of zero
c/a =
-4/3 = 4/3
2. Find a
quadratic polynomial each with the given numbers as the sum and product of its
zeroes respectively.
(i)1/4 , -1 (ii) √2 , 1/3 (iii) 0,
√5 (iv) 1,1 (v) -1/4 ,1/4 (vi) 4,1
Now formula
of quadratic equation is
Plug the value in formula
we get
X^2 –(1/4)x -1 = 0
Multiply by 4 to remove
denominator we get
4x^2 - x -4 = 0
(ii) √2 , 1/3
Now formula of quadratic equation is
Plug the value in formula
we get
x^2 –(√2)x +
1/3 = 0
Multiply by 3 to remove
denominator we get
3x^2 - 3√2 x +
1 = 0
(iii) 0, √5
Now formula
of quadratic equation is
Plug the value in formula
we get
x^2 –(0)x + √5 = 0
simplify it we get
x^2 + √5 = 0
(iv) 1,1
Now formula
of quadratic equation is
Plug the value in formula
we get
x^2 –(1)x +
1 = 0
simplify it we get
x^2 - x +
1 = 0
(v) -1/4
,1/4
Now formula
of quadratic equation is
Plug the value in formula
we get
x^2 –(-1/4)x +
1/4 = 0
multiply by 4 we get
4x^2 + x +
1 = 0
(vi) 4,1
Now formula
of quadratic equation is
Plug the value in formula
we get
x^2 –(4)x + 1 = 0
x^2 –4x + 1 = 0
EXERCISE 2.3
1. Divide
the polynomial p(x) by the polynomial g(x) and find the quotient and remainder
in each of
the following :
(i)
p(x)
= x3 – 3x2 + 5x – 3, g(x) = x2 – 2
So quotient
= x-3 and remainder 7x - 9
(ii)
p(x)
= x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
So quotient = x^2 +x-3 and remainder 8
(iii)
p(x)
= x4 – 5x + 6, g(x) = 2 – x2
2. Check
whether the first polynomial is a factor of the second polynomial by dividing
the
second
polynomial by the first polynomial:
(i)t2
– 3, 2t4 + 3t3 – 2t2 – 9t – 12
Remainder
is 0 hence t^2 -3 is a factor of 2t^4
+ 3t^3 – 2t^2 – 9t – 12
(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
(3) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Remainder
is 2 so x^3 – 3x + 1 is
not a factor
3. Obtain
all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if
two of its zeroes are √(5/3) and -
√(5/3)
Quotient is
Compare the
equation with ax^2 + bx +
c = 0
We get
a = 1 ,b=2
c= 1
To
factorize the value we have to find two value which
sum is
equal to b = 2
product
is a*c =
1*1 = 1
1 and 1 are such number which
sum is 1 + 1 = 2
product is 1 * 1 = 1
So we can
write middle term 2x = x +
x
We get
x (x +1)
+1(x +1) = 0
( x + 1)( x
+ 1) = 0
x+1 = 0 ,
x+1 = 0
x=
-1 , x= - 1
so our zeroes are
- 1 -1
, √(5/3) and - √(5/3)
4. On
dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x – 2
and –2x + 4, respectively. Find g(x).
according
to division algorithm
Dividend = Divisor × Quotient + Remainder
p(x) =
g(x) × q(x) +
r(x),
plug the
value in formula we get
x^3
– 3x^2 + x + 2
= g(x) *(x-2) - 2x + 4
Add 2x and
subtract 4 both side we get
x^3
– 3x^2 + x + 2 + 2x – 4 = g(x) *(x-2)
simplify
and divide by x/2 we get
(x^3
– 3x^2 + 3x– 2)/(x-2) = g(x)
So we
get g(x) = x^2 – x +1
5. Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm
And
We can
write many such examples
deg p(x) = deg q(x)
We can
write many such examples
P(x) = x^2
, q(x) = x^2 g(x) = 1 R(x) = 0
(ii) deg q(x) = deg r(x)
Q(x) = x ,
R(x) = x , p(x) = x^3 + x g(x) = x^2
(iii) deg r(x) = 0
Q(x) = 1 , R(x) = 1 ,
p(x) = x+1, g(x) = x