Polynomials(Poly= many, Nomials=Terms)
If a
function f(x) is polynomial, highest power of x is called Degree of polynomial.
For eg
f(x)=ax+b
can be said f(x) is polynomial of degree 1/ linear polynomial,
f(x)=ax2+bx+c
can be said f(x) is polynomial of degree 2/quadratic polynomial,
f(x)=ax3+bx2+cx+d
can be said f(x) is polynomial of degree 3/ Cubic Polynomial,
Linear
polynomials when drawn between x & y axis forms a straight line.
Quadratic polynomials when
drawn between x & y axis form parabolas or
These
parabolas can intersect x axis at 2 points, 1 point or no points
No. of
zeros means value of Y is 0 at that point. So
0
1
3
2
4
3
Assume
α & β are
zeros of equation f(x)=ax2+bx+c
ax2+bx+c= k(x- α)(x- β)
=k(x2-
αx- βx+α β)
=k(x2-(α+β)x+ α β)
ax2+bx+c
=kx2-k(α+β)x+k α β
a=k,
b=-k(α+β), c=kαβ
it
means –b/a= α+β & c/a= αβ
For
cubic equation of ax3+bx2+cx+d which has αβƔ as
zeros
–b/a= α+β+Æ” ; c/a=( αβ + βƔ + αƔ) & -d/a= αβƔ
Excersize
2.2
Answer 1 i) x2-2x-8=0 => x2-4x+2x-8
=>x(x-4)+2(x-4)=(x-4)(x+2)=0 so zeros are x=4, x=-2 (4,-2) sum of zeros
means –b/a=
α+β &
c/a= αβ sum of zeros=–(-2)/1=2
product
of zeros=-8/1=-8
ii)4s2-4s+1=0,
4s2-2s-2s+1=2s(2s-1)-1(2s-1)=(2s-1)(2s-1)=0 s=1/2.1/2 hence zeros are (1/2.1/2)
sum of
zeros==-b/a= -(-4/4)=1= α+β, product of zeros=c/a=1/4= αβ
iii)6x2-3-7x = 6x2-7x-3
6*3=18,
(9*2)è 6x2-9x+2x-3=3x(2x-3)+1(2x-3)=(2x-3)(3x+1)
X=3/2,-1/3 sum of zeros 3/2-1/3= (3*3-1*2)/6=7/6=-b/a
Product
3/2*-1/3=-1/2
iv)4u2+8u=0, 4u(u+2)
so
zeros are u=0,-2 sum of zeros=-b/a = -8/4=-2
product
of zeros = c/a=0/4=0
v)t2-15=0
t=+- √15
sum of
zeros =+ √15- √15=0=-b/a
product
of zeros=c/a=-15/1=-15= √15*- √15=-15
vi)3x2-x-4=0
3x2-4x+3x-4=x(3x-4)+1(3x-4)=(3x-4)(x+1)=x=4/3,x=-1
sum
4/3-1=1/3= -b/a, product =c/a=4/3*-1=-4/3
2.i) From the formulas of sum
and product of zeroes, we know,
Sum of
zeroes = α+β
Product
of zeroes = α β
Sum of
zeroes = α+β = 1/4
Product
of zeroes = α β = -1
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic
polynomial equation can be written directly
as:-
x2–(α+β)x +αβ = 0
x2–(1/4)x
+(-1) = 0
4x2–x-4
= 0
Thus,4x2–x–4
is the quadratic polynomial.
(ii)√2, 1/3
Solution:
Sum of
zeroes = α + β =√2
Product
of zeroes = α β = 1/3
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic
polynomial equation can be written directly
as:-
x2–(α+β)x +αβ = 0
x2 –(√2)x + (1/3) = 0
3x2-3√2x+1 = 0
Thus,
3x2-3√2x+1 is the quadratic polynomial.
(iii) 0, √5
Solution:
Given,
Sum of
zeroes = α+β = 0
Product
of zeroes = α β = √5
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic
polynomial equation can be written directly
as:-
x2–(α+β)x +αβ = 0
x2–(0)x
+√5= 0
Thus,
x2+√5 is the quadratic polynomial.
(iv) 1, 1
Solution:
Given,
Sum of
zeroes = α+β = 1
Product
of zeroes = α β = 1
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic
polynomial equation can be written directly
as:-
x2–(α+β)x +αβ = 0
x2–x+1
= 0
Thus ,
x2–x+1is the quadratic polynomial.
(v) -1/4, 1/4
Solution:
Given,
Sum of
zeroes = α+β = -1/4
Product
of zeroes = α β = 1/4
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic polynomial
equation can be written directly
as:-
x2–(α+β)x +αβ = 0
x2–(-1/4)x
+(1/4) = 0
4x2+x+1
= 0
Thus,4x2+x+1
is the quadratic polynomial.
(vi) 4, 1
Solution:
Given,
Sum of
zeroes = α+β = 4
Product
of zeroes = αβ = 1
∴ If α and β are zeroes of any quadratic polynomial, then the quadratic
polynomial equation can be written directly
as:-
x2–(α+β)x+αβ = 0
x2–4x+1
= 0
Thus,x2–4x+1
is the quadratic polynomial.
Excercize 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
Solution:
Given,
Dividend
= p(x) = x3-3x2+5x–3
Divisor
= g(x) = x2– 2
Therefore,
upon division we get,
Quotient
= x–3
Remainder
= 7x–9
(ii)
p(x) = x4-3x2+4x+5 , g(x) = x2+1-x
Solution:
Given,
Dividend
= p(x) = x4 - 3x2 + 4x +5
Divisor
= g(x) = x2 +1-x
Therefore,
upon division we get,
Quotient
= x2 + x–3
Remainder
= 8
(iii)
p(x) =x4–5x+6, g(x) = 2–x2
Solution:
Given,
Dividend
= p(x) =x4 - 5x + 6 = x4 +0x2–5x+6
Divisor
= g(x) = 2–x2 = –x2+2
Therefore,
upon division we get,
Quotient
= -x2-2
Remainder
= -5x + 10
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
Solutions:
Given,
First
polynomial = t2-3
Second
polynomial = 2t4 +3t3-2t2 -9t-12
As we
can see, the remainder is left as 0. Therefore, we say that, t2-3 is a factor
of 2t2+3t+4.
(ii)x2+3x+1
, 3x4+5x3-7x2+2x+2
Solutions:
Given,
First
polynomial = x2+3x+1
Second
polynomial = 3x4+5x3-7x2+2x+2
As we
can see, the remainder is left as 0. Therefore, we say that, x2 + 3x + 1 is a
factor of 3x4+5x3-7x2+2x+2.
(iii)
x3-3x+1, x5-4x3+x2+3x+1
Solutions:
Given,
First
polynomial = x3-3x+1
Second
polynomial = x5-4x3+x2+3x+1
As we
can see, the remainder is not equal to 0. Therefore, we say that, x3-3x+1 is
not a factor of x5-4x3+x2+3x+1 .
3.
Obtain all other zeroes of 3x4+6x3-2x2-10x-5, if two of its zeroes are √(5/3) and - √(5/3).
Solutions:
Since
this is a polynomial equation of degree 4, hence there will be total 4 roots.
√(5/3) and - √(5/3) are zeroes of polynomial f(x).
∴ (x -√(5/3)) (x+√(5/3) = x2-(5/3) = 0
(3x2−5)=0, is a factor of given polynomial f(x).
Now,
when we will divide f(x) by (3x2−5) the quotient obtained will also be a factor
of f(x) and the remainder
will
be 0.
Therefore,
3x4 +6x3 −2x2 −10x–5 = (3x2 –5)(x2+2x+1)
Now,
on further factorizing (x2+2x+1) we get,
x2+2x+1
= x2+x+x+1 = 0
x(x+1)+1(x+1)
= 0
(x+1)(x+1)
= 0
So,
its zeroes are given by: x= −1 and x = −1.
Therefore,
all four zeroes of given polynomial equation are:
√(5/3),- √(5/3) , −1 and −1.
Hence,
is the answer.
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).
Solutions:
Given,
Dividend,
p(x) = x3-3x2+x+2
Quotient
= x-2
Remainder
= –2x+4
We
have to find the value of Divisor, g(x) =?
As we
know,
Dividend
= Divisor × Quotient + Remainder
∴ x3-3x2+x+2 = g(x)×(x-2) + (-2x+4)
x3-3x2+x+2-(-2x+4)
= g(x)×(x-2)
Therefore,
g(x) × (x-2) = x3-3x2+x+2
Now,
for finding g(x) we will divide x3-3x2+x+2 with (x-2)
Therefore,
g(x) = (x2–x+1)
5. Give examples of polynomials p(x), g(x), q(x) and r(x), which
satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0
Solutions:
According
to the division algorithm, dividend p(x) and divisor g(x) are two polynomials,
where g(x)≠0. Then we
can
find the value of quotient q(x) and remainder r(x), with the help of below
given formula;
Dividend
= Divisor × Quotient + Remainder
∴ p(x) = g(x)×q(x)+r(x)
Where
r(x) = 0 or degree of r(x)< degree of g(x).
Now
let us proof the three given cases as per division algorithm by taking examples
for each.
(i) deg p(x) = deg q(x)
Degree
of dividend is equal to degree of quotient, only when the divisor is a constant
term.
Let us
take an example, 3x2+3x+3 is a polynomial to be divided by 3.
So,
(3x2+3x+3)/3 = x2+x+1 = q(x)
Thus,
you can see, the degree of quotient is equal to the degree of dividend.
Hence,
division algorithm is satisfied here.
(ii) deg q(x) = deg r(x)
Let us
take an example , p(x)=x2+x is a polynomial to be divided by g(x)=x.
So, (x2+x)/x
= x+1 = q(x)
Also,
remainder, r(x) = 0
Thus,
you can see, the degree of quotient is equal to the degree of remainder.
Hence,
division algorithm is satisfied here.
(iii) deg r(x) = 0
The
degree of remainder is 0 only when the remainder left after division algorithm
is constant.
Let us
take an example, p(x) = x2+1 is a polynomial to be divided by g(x)=x.
So,( x2+1)/x=
x=q(x)
And
r(x)=1
Clearly,
the degree of remainder here is 0.
Hence,
division algorithm is satisfied here.
Exercise 2.4
1. Verify that the numbers given alongside of the cubic
polynomials below are their zeroes. Also verify the relationship between the
zeroes and the coefficients in each case:
(i) 2x3+x2-5x+2; -1/2, 1, -2
Solution:
Given,
p(x) = 2x3+x2-5x+2
And
zeroes for p(x) are = 1/2, 1, -2
∴ p(1/2) = 2(1/2)3+(1/2)2-5(1/2)+2 = (1/4)+(1/4)-(5/2)+2 = 0
p(1) =
2(1)3+(1)2-5(1)+2 = 0
p(-2)
= 2(-2)3+(-2)2-5(-2)+2 = 0
Hence,
proved 1/2, 1, -2 are the zeroes of 2x3+x2-5x+2.
Now,
comparing the given polynomial with general expression, we get;
∴ ax3+bx2+cx+d = 2x3+x2-5x+2
a=2,
b=1, c= -5 and d = 2
As we know,
if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;
α +β+γ = –b/a
αβ+βγ+γα = c/a
α βγ = – d/a.
Therefore,
putting the values of zeroes of the polynomial,
α+β+γ = ½+1+(-2) = -1/2 = –b/a
αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) = -5/2 = c/a
α β γ = ½×1×(-2) = -2/2 = -d/a
Hence,
the relationship between the zeroes and the coefficients are satisfied.
(ii) x3-4x2+5x-2 ;2, 1, 1
Solution:
Given,
p(x) = x3-4x2+5x-2
And
zeroes for p(x) are 2,1,1.
∴ p(2)= 23-4(2)2+5(2)-2 = 0
p(1) =
13-(4×12 )+(5×1)-2 = 0
Hence
proved, 2, 1, 1 are the zeroes of x3-4x2+5x-2
Now,
comparing the given polynomial with general expression, we get;
∴ ax3+bx2+cx+d = x3-4x2+5x-2
a = 1,
b = -4, c = 5 and d = -2
As we
know, if α, β, γ are the zeroes of the cubic polynomial ax3+bx2+cx+d , then;
α + β + γ = –b/a
αβ + βγ + γα = c/a
α β γ = – d/a.
Therefore,
putting the values of zeroes of the polynomial,
α +β+γ = 2+1+1 = 4 = -(-4)/1 = –b/a
αβ+βγ+γα = 2×1+1×1+1×2 = 5 = 5/1= c/a
αβγ = 2×1×1 = 2 = -(-2)/1 = -d/a
Hence,
the relationship between the zeroes and the coefficients are satisfied.
2. Find a cubic polynomial with the sum, sum of the product of its
zeroes taken two at a time, and the
product of its zeroes as 2, –7, –14 respectively.
Solution:
Let us
consider the cubic polynomial is ax3+bx2+cx+d and the values of the zeroes of
the polynomials be α, β, γ.
As per
the given question,
α+β+γ = -b/a = 2/1
αβ +βγ+γα = c/a = -7/1
α βγ = -d/a = -14/1
Thus,
from above three expressions we get the values of coefficient of polynomial.
a = 1,
b = -2, c = -7, d = 14
Hence,
the cubic polynomial is x3-2x2-7x+14
3. If the zeroes of the polynomial x3-3x2+x+1 are a – b, a, a + b, find a and b.
Solution:
We are
given with the polynomial here,
p(x) =
x3-3x2+x+1
And
zeroes are given as a – b, a, a + b
Now,
comparing the given polynomial with general expression, we get;
∴px3+qx2+rx+s = x3-3x2+x+1
p = 1,
q = -3, r = 1 and s = 1
Sum of
zeroes = a – b + a + a + b
-q/p =
3a
Putting
the values q and p.
-(-3)/1
= 3a
a=1
Thus,
the zeroes are 1-b, 1, 1+b.
Now,
product of zeroes = 1(1-b)(1+b)
-s/p =
1-b2
-1/1 =
1-b2
b2 =
1+1 = 2
b = √2
Hence,1-√2, 1 ,1+√2 are the zeroes of x3-3x2+x+1.
4. If two zeroes of the polynomial x4-6x3-26x2+138x-35 are 2 ±√3, find other zeroes.
Solution:
Since
this is a polynomial equation of degree 4, hence there will be total 4 roots.
Let
f(x) = x4-6x3-26x2+138x-35
Since
2 +√3 and 2-√3 are zeroes of given polynomial f(x).
∴ [x−(2+√3)] [x−(2-√3)] = 0
(x−2−√3)(x−2+√3) = 0
On
multiplying the above equation, we get,
x2-4x+1,
this is a factor of a given polynomial f(x).
Now,
if we will divide f(x) by g(x), the quotient will also be a factor of f(x) and
the remainder will be 0.
So, x4-6x3-26x2+138x-35
= (x2-4x+1)(x2 –2x−35)
Now,
on further factorizing (x2–2x−35) we get,
x2–(7−5)x −35 = x2– 7x+5x+35 = 0
x(x −7)+5(x−7) = 0
(x+5)(x−7) = 0
So,
its zeroes are given by:
x= −5 and x = 7.
Therefore,
all four zeroes of given polynomial equation are: 2+√3 , 2-√3, −5 and 7
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