Important Key points

Example 3.30
Show that the polynomial$9x^9 + 2x^5 − x^4 − 7x^2 + 2$ has at least six imaginary roots.

Solution :
Let p(x) = $9x^9 + 2x^5 − x^4 − 7x^2 + 2$
Signs of p(x) is +, +, -, -, +
No.of sign Changes in coefficients of p(x) is 2 (+$\rightarrow$-, -$\rightarrow$+)
$\Rightarrow$ Number of positive roots of P( x) cannot be more than two.

Finding p(-x) = $9(-x)^9 + 2(-x)^5 − (-x)^4 − 7(-x)^2 + 2$
$\hspace{1.7cm}$$= -9x^9 - 2x^5 − x^4 − 7x^2 + 2$
Signs of p(-x) is -, -, -, -, +
No.of sign Changes in coefficients of p(-x) is 1 (-$\rightarrow$+)
$\Rightarrow$The number of negative roots cannot be more than one.

Checking whether '0' is root of p(x).
p(0)=0 +0 + 0+ 0+ 2 = 2
$\Rightarrow$ '0' is not a root.

To find the lower bound for the number of imaginary roots of p(x) $$atleast\ no.\ of\ imaginary\ roots = n-(m+k)$$ where
n=Degree of p(x)
m = the number of sign changes in coefficients of P(x),
k = the number of sign changes in coefficients of P(-x)
Here n=9, m=2, n=1
$\Rightarrow$ atleast no. of imaginary roots = 9-(2+1) = 6
So maximum number of real roots is 3 and hence there are atleast six imaginary roots.

Example 3.31
Discuss the nature of the roots of the following polynomials:
(i) $x^{2018} +1947x^{1950} +15x^8 + 26x^6 + 2019$
(ii) $x^5 −19x^4 + 2x^3 + 5x^2 +11$

Solution : P(x) be the polynomial under consideration.
(i) Let p(x)= $x^{2018} +1947x^{1950} +15x^8 + 26x^6 + 2019$
Signs of p(x) is +, +, +, +, +
No.of sign Changes in coefficients of p(x) is 0
$\Rightarrow$ Number of positive roots of P( x) is zero. i.e, No positive roots

Finding p(-x)= $(-x)^{2018} +1947(-x)^{1950} +15(-x)^8 + 26(-x)^6 + 2019$
$\hspace{1.7cm}$ = $x^{2018} +1947x^{1950} +15x^8 + 26x^6 + 2019$
Signs of p(x) is +, +, +, +, +
No.of sign Changes in coefficients of p(x) is 0
$\Rightarrow$The number of negative roots is zero. i.e, No negative roots.

Checking whether '0' is root of p(x).
p(0)=0 +0 + 0+ 0+ 2019 = 2019
$\Rightarrow$ '0' is not a root.
Thus the polynomial has no real roots and hence all roots of the polynomial are imaginary roots..

(ii) Let p(x) = $x^5 −19x^4 + 2x^3 + 5x^2 +11$
Signs of p(x) is +, -, +, +, +
No.of sign Changes in coefficients of p(x) is 2 (+$\rightarrow$-, -$\rightarrow$+)
$\Rightarrow$ Number of positive roots of P( x) cannot be more than two.

Finding p(-x) $= (-x)^5 −19(-x)^4 + 2(-x)^3 + 5(-x)^2 +11$
$\hspace{1.7cm}$ = $-x^5 -19x^4 - 2x^3 + 5x^2 +11$
Signs of p(-x) is -, -, -, +, +
No.of sign Changes in coefficients of p(-x) is 1 (-$\rightarrow$+)
$\Rightarrow$The number of negative roots cannot be more than one.

Checking whether '0' is root of p(x).
p(0)=0 +0 + 0+ 0+ 11 = 11
$\Rightarrow$ '0' is not a root.

To find the lower bound for the number of imaginary roots of p(x) $$atleast\ no.\ of\ imaginary\ roots = n-(m+k)$$ where
n=Degree of p(x)
m = the number of sign changes in coefficients of P(x),
k = the number of sign changes in coefficients of P(-x)
Here n=5, m=2, n=1
$\Rightarrow$ atleast no. of imaginary roots = 5-(2+1) = 2
So maximum number of real roots is 3 and hence there are atleast two imaginary roots.

Exercise 3.6

1. Discuss the maximum possible number of positive and negative roots of the polynomial equation $9x^9 – 4x^8 + 4x^7 – 3x^6 + 2x5^ + x^3 + 7x^2 + 7^x + 2 = 0$.

Solution :
Let p(x)= $9x^9 – 4x^8 + 4x^7 – 3x^6 + 2x^5 + x^3 + 7x^2 + 7x + 2 = 0$.
Signs of p(x) is +, -, +, -, +, +, +, +, +,
No.of sign Changes in coefficients of p(x) is 4 (+$\rightarrow$-, -$\rightarrow$+, +$\rightarrow$-, -$\rightarrow$+)
$\Rightarrow$ Number of positive roots of P( x) cannot be more than Four.

Finding p(-x)= $9(-x)^9 – 4(-x)^8 + 4(-x)^7 – 3(-x)^6 + 2(-x)^5 + (-x)^3 + 7(-x)^2 + 7(-x) + 2 = 0$.
$\hspace{1.7cm}$ = $-9x^9 – 4x^8 - 4x^7 – 3x^6 - 2x^5 - x^3 + 7x^2 - 7^x + 2 = 0$.
Signs of p(-x) is -, -, -, -, -, -, +, -, +
No.of sign Changes in coefficients of p(-x) is 2 (+$\rightarrow$-, -$\rightarrow$+)
$\Rightarrow$The number of negative roots cannot be more than two.

Checking whether '0' is root of p(x).
p(0)=0 +0 + 0+ 0+ 0+ 0+ 0+ 0+ 7 = 7
$\Rightarrow$ '0' is not a root.

So maximum number of real roots is 6

2. Discuss the maximum possible number of positive and negative zeros of the polynomials $x^2$ – 5x + 6 and $x^2$ – 5x + 16. Also, draw a rough sketch of the graphs.

Solution :
Let p(x) = $x^2$ – 5x + 6
Signs of p(x) is +, -, +
No.of sign Changes in coefficients of p(x) is 2 (+$\rightarrow$-, -$\rightarrow$+)
$\Rightarrow$ Number of positive roots of P( x) cannot be more than two.

Finding p(-x)= $(-x)^2 – 5(-x) + 6$.
$\hspace{1.7cm}$ = $x^2 + 5x + 6$.
Signs of p(-x) is +, +, +
No.of sign Changes in coefficients of p(-x) is 0
$\Rightarrow$The number of negative roots cannot be more than zero. i.e, No negative roots.

3. Show that the equation $x^9 – 5x^5 + 4x^4 + 2x^2 +1 = 0$ has atleast 6 imaginary solutions.

Solutions :
Let p(x) = $x^9 – 5x^5 + 4x^4 + 2x^2 +1 = 0$
Signs of p(x) is +, -, +, +, +,
No.of sign Changes in coefficients of p(x) is 2 (+$\rightarrow$-, -$\rightarrow$+)
$\Rightarrow$ Number of positive roots of P( x) cannot be more than two.

Finding p(-x)= $(-x)^9 – 5(-x)^5 + 4(-x)^4 + 2(-x)^2 +1 = 0$.
$\hspace{1.7cm}$ = $-x^9 + 5x^5 + 4x^4 + 2x^2 +1 = 0$
Signs of p(-x) is -, +, +, +, +
No.of sign Changes in coefficients of p(-x) is 1 ( -$\rightarrow$+)
$\Rightarrow$The number of negative roots cannot be more than one.

Checking whether '0' is root of p(x).
p(0)=0 +0 + 0+ 0+ 1 = 1
$\Rightarrow$ '0' is not a root.

To find the lower bound for the number of imaginary roots of p(x) $$atleast\ no.\ of\ imaginary\ roots = n-(m+k)$$ where
n=Degree of p(x)
m = the number of sign changes in coefficients of P(x),
k = the number of sign changes in coefficients of P(-x)
Here n=9, m=2, n=1
$\Rightarrow$ atleast no. of imaginary roots = 9-(2+1) = 6
So maximum number of real roots is 3 and hence there are atleast six imaginary roots.

4. Determine the number of positive and negative roots of the equation $x^9 – 5x^8 – 14x^7 = 0$

Solution :
Let p(x)= $x^9 – 5x^8 – 14x^7 = 0$
Signs of p(x) is +, -, -
No.of sign Changes in coefficients of p(x) is 1 (+$\rightarrow$-)
$\Rightarrow$ Number of positive roots of P( x) cannot be more than one.

Finding p(-x)= $(-x)^9 – 5(-x)^8 – 14(-x)^7 = 0$
$\hspace{1.7cm}$ = $-x^9 – 5x^8 + 14x^7 = 0$
Signs of p(-x) is -, -, +
No.of sign Changes in coefficients of p(-x) is 1 ( -$\rightarrow$+)
$\Rightarrow$The number of negative roots cannot be more than one.

Checking whether '0' is root of p(x).
p(0)=0 +0 + 0 = 0
$\Rightarrow$ '0' is a root.
So there are 3 real roots.

To find the lower bound for the number of imaginary roots of p(x) $$atleast\ no.\ of\ imaginary\ roots = n-(m+k)$$ where
n=Degree of p(x)
m = the number of sign changes in coefficients of P(x),
k = the number of sign changes in coefficients of P(-x)
Here n=9, m=2, n=1
$\Rightarrow$ atleast no. of imaginary roots = 9-(3) = 6
So maximum number of real roots is 3 and hence there are atleast six imaginary roots.

5. Find the exact number of real zeros and imaginary of the polynomial $x^9 + 9x^7 + 7x^5 + 5x^3 + 3x.$

Solution :
Let p(x)= $x^9 + 9x^7 + 7x^5 + 5x^3 + 3x.$
Signs of p(x) is +, +, +, +, +,
No.of sign Changes in coefficients of p(x) is 0.
$\Rightarrow$ Number of positive roots of P( x) cannot be more than zero. i.e, No positive roots

Finding p(-x)= $(-x)^9 + 9(-x)^7 + 7(-x)^5 + 5(-x)^3 + 3(-x).$
$\hspace{1.7cm}$ = $-x^9 - 9x^7 - 7x^5 - 5x^3 - 3x.$
Signs of p(-x) is -, -, -, -, -
No.of sign Changes in coefficients of p(-x) is 0
$\Rightarrow$The number of negative roots cannot be more than zero.i.e, No negative roots
It has no positive real roots and no negative real roots.

Checking whether '0' is root of p(x).
p(0)=0 +0 + 0+ 0+ 0 = 0
$\Rightarrow$ '0' is a root.
No.of real root is One.

To find the lower bound for the number of imaginary roots of p(x) $$atleast\ no.\ of\ imaginary\ roots = n-(m+k)$$ where
n=Degree of p(x)
m = the number of sign changes in coefficients of P(x),
k = the number of sign changes in coefficients of P(-x)
Here n=9, m=2, n=1
$\Rightarrow$ atleast no. of imaginary roots = 9-(1) = 8
So maximum number of real roots is 1 and since degree of polynomial is 9 and hence there are atleast eight imaginary roots.