WBBSE Solutions For Class 9 Maths Chapter 7 Polynomial

Class 9 Math Solution WBBSE Chapter 7 Polynomial Exercise 7

Question 1. If f(x) = x⁵ + 3x³ – 7x² + 6,h(x)=3x³-8x² + 7,g(x) = x + 1,p(x) = x⁴ – x² + 2 and q (y) = 7y³-y+ 10 then let us calculate and write what would be the following polynomials:

1. f(x)+g(x)

Solution: f(x)+g(x) =x⁵+3x³-7x²+6+x+1 =x⁵+3x³-7x²+x+7

2. f(x)-h(x)

Solution: f(x)-h(x) =x⁵+3x³-7x²+6-(3x³-8x²+7) =x⁵+3x³-7x²+6-3x³+8x²-7 = x⁵+x²-1

3. f(x) -p(x)

Solution: f(x)-p(x) =x⁵+3x³-7x²+6-(x⁴-x²+2) =x⁵+3x³-7x²+6-x⁴+x²-2 =x⁵-x⁴+3x³-6x²+4

4. f(x)+p(x)

Answer: f(x)+p(x) =x⁵+3x³-7x²+6+x⁴-x²+2 =x⁵+x⁴+3x³-8x²+8

5. p(x)+g(x)+f(x)

Solution: p(x)+g(x)+f(x) = x⁴-x²+2+x+1+x⁵+3x³-7x²+6 =x⁵+x⁴-3x³-8x²+x+7

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6. p(x)-q(y)

Solution: p(x)-q(y) =x⁴-x²+2-(7y³-y+10) =x⁴-x²+2-7y³+y-10 =x⁴-7y³-x²+y-8

7. f(x).g(x)

Solution: f(x).g(x) = (x⁵+3x³-7x²+6)(x+1) =x⁶+3x⁴-7x³+6x+x⁵+3x³-7x²+6 =x+x⁵+3x⁴-4x³-7x²+6x+6

8. p(x).g(x)

Solution: p(x).g(x) = (x⁴-x²+2)(x+1) =x⁵-x³+2x+x⁴-x²+2 = x⁵+x⁴-x³-x²+2x+2

Ganit Prakash Class 9 Solutions Chapter 7 Polynomial Exercise 7.1

Question 1. Let us write which are the polynomials in the following algebraic expressions. Let us write the degree of each of the polynomials.

1. 2x⁶ – 4x⁵ +7x² +3

Solution: 2x⁶ – 4x⁵ +7x² +3 is a polynomial because the index of the variable is a whole number and the highest index of x is 6. So the degree of 2x⁶ – 4x⁵ +7x² +3 is 6.

2. x² + 2x^-1 +4

Solution: x² + 2x^-1 +4 is not a polynomial as the index of the variable is not a whole number.

3. \(y^3-\frac{3}{4} y+\sqrt{7}\)

Solution: \(y^3-\frac{3}{4} y+\sqrt{7}\) is a polynomial as the index of the variable is a whole number, and the highest index of y is 3, so the degree of \(y^3-\frac{3}{4} y+\sqrt{7}\) is 3.

4. \(\frac{1}{x}-x+2\)

Solution: \(\frac{1}{x}-x+2\)  is not polynomial as the index of the variable is not a whole number.

5. \(x^{51}-1\)

Solution: \(x^{51}-1\) is a polynomial as the index of the variable is a whole number, and the highest index of x is 51 so the degree of \(x^{51}-1\)  is 51.

6. \(3 \sqrt{t}+\frac{t}{27}\)

Solution: \(3 \sqrt{t}+\frac{t}{27}=t^{\frac{1}{3}}+\frac{t}{27}\) is not a polynomial as the index of the variable is not a whole number.

7. 15

Solution: 15 15.1 15.x° It is a monomial of degree 0.

8. 0

Solution: 0 Power of the polynomial is undefined.

9. \(z+\frac{3}{z}+2\)

Solution: \(z+\frac{3}{z}+2=z+3 z^{-1}+2\) is not a polynomial as the index of the variable is not a whole number.

10. y³+ 4

Solution: y3+4 is not a polynomial as the index of the variable is a whole number. So the degree of y³+ 4 is 3.

11. \(\frac{1}{\sqrt{2}} x^2-\sqrt{2} x+2\)

Solution: \(\frac{1}{\sqrt{2}} x^2-\sqrt{2} x+2\) is not a polynomial as the index of the variable is a whole number. So the degree of \(\frac{1}{\sqrt{2}} x^2-\sqrt{2} x+2\) is 2.

Question 2. In the following polynomials, let us write which are first-degree polynomials in one variable, which are second-degree polynomials in one variable, and which are third-degree polynomials in one variable.

1. 2x + 17

Solution: 2x + 17 One variable – Degree – 1

2. x³ + x² + x + 1

Solution: x³ + x² + x + 1 One variable – Degree – 3

3. – 3+2y²+5xy

Solution: – 3+2y²+5xy is Not one variable here; x & y are two variables.

4. 5-x-x³

Solution: 5-x-x³ One variable – Degree – 3

5. √2+t- t²

Solution: √2+t- t² One variable – Degree – 2

6. √5x

Solution: √5x One variable – Degree – 1

Class 9 Mathematics West Bengal Board

Question 3. Let us write the coefficients of the following polynomials according to the guidelines:

1. The co-efficient of x³ in 5x³ – 13x² + 2

Solution: In 5x³ – 13x² + 2, co-efficient of x3 is 5.

2. The coefficient of x in x²-x+2.

Solution: In x²-x+2, the co-efficient of x is 1.

3. The co-efficient of x² in 8x-19

Solution: In 8x-19= 0x² + 8x-19, co-efficient of x² is 0.

4. The co-efficient of x° in √11-3√11x+x²

Solution: In √11-3√11x+x² = √11x° -3√11x+x², co-efficient of x° is √11.

Question 4. I write the degree of each of the following polynomials :

1. x⁴ + 2x³ +x² + x

Solution: x⁴ + 2x³ +x² + x Degree – 4

2. 7x-5

Solution: 7×5 Degree – 1

3. 16

Solution: 16 = 16.1 16x° Degree – 0

4. 2-y-y³

Solution: 2-y-y³ Degree – 3

5. 7t

Solution: 7t Degree – 1

6. 5 – x² + x¹⁹

Solution: 5 – x² + x¹⁹ Degree – 19

Class 9 Mathematics West Bengal Board

Question 5. I write two separate binomials in one variable whose degrees are 17.

1. 5x¹⁷+1

Solution: 5x¹⁷+1 Binomial with two variables & degree 14.

2. 2y¹⁷-5

Solution: 2y¹⁷-5 Binomial with two variables & degree 17.

Question 6. I write two separate monomials in one variable whose degrees are 4.

1. 2x⁴

Solution: 2x⁴ Monomial with one variable & degree 4.

2. 3y⁴

Solution:: 3y⁴ Monomial with one variable & degree 4.

Question 7. I write two separate trinomials in one variable whose degrees are 3.

1.  2x³ + 3x² + 4x

Solution: 2x³ + 3x² + 4x Trinomial with one variable & degree 3.

2. y³+2y²+5

Solution: y³+2y²+5 Trinomial with one variable & degree 3.

Class 9 Math Solution WBBSE

Question 8. In the following algebraic expressions, which are polynomials in one variable, which are polynomials in two variables, and which are not polynomials – Let us write them.

1.  x² + 3x + 2

Solution: x² + 3x + 2, one variable

2. x² + y² + a²

Solution: x² + y² + a², one variable

3. y²– 4ax

Solution: y² – 4ax, one variable

4. x + y + 2

Solution: x + y +2, one variable

5. x⁸+y⁴+ x⁵y⁹

Solution: x⁸+y⁴+ x⁵y⁹, one variable

6. \(x+\frac{5}{x}\)

Solution: \(x+\frac{5}{x}=x+5 x^{-1}\) is not a polynomial as the degree of variable is not a whole number.

WBBSE Class 9 Maths Solutions Chapter 7 Polynomial Exercise 7.2

Question 1. If f(x) = x²+9x-6, then let us write by calculating the values of f(0), f(1) and f(3)

Solution:

Given

f(x) = x²+9x-6

\(\begin{aligned}
&f(0)=(0)^2+9.0-6=-6 \\
& f(1)=(1)^2+9.1-6=10-6=4 \\
& f(3)=(3)^2+9.3-6=9+27-6 \\
& =36-6=30 \\
&
\end{aligned}\)

Question 2. By calculating the following polynomials f(x), let us write the values of f(1) and f(-1).

1. f(x) = 2x³ + x² + x + 4

Solution:

Given

f(x) = 2x³ + x² + x + 4

\(\begin{aligned}
f(1) & =2(1)^3+(1)^2+1+4 \\
& =2+1+1+4=8 \\
f(-1) & =2(-1)^3+(-1)^2+(-1)+4 \\
& =-2+1-1+4 \\
& =5-3 \\
& =2
\end{aligned}\)

2. f(x) = 3x⁴– 5x³ + x² + 8

Solution:

Given

f(x) = 3x⁴– 5x³ + x² + 88

\(\begin{aligned}
f(1) & =3(1)^4-5(1)^3+(1)^2+8 \\
& =3-5+1+8 \\
& =7 \\
f(-1) & =3(-1)^4-5(-1)^3+(-1)^2+8 \\
& =3+5+1+8 \\
& =17
\end{aligned}\)

3. f(x) = 4 + 3x – x³ + 5x⁶

Solution:

Given

f(x) = 4 + 3x – x³ + 5x⁶

\(\begin{aligned}
f(1) & =4+3.1-(1)^3+5(1)^6 \\
& =4+3-1+5 \\
& =11 \\
f(-1) & =4+3(-1)-(-1)^3+5(-1)^6 \\
& =4-3+1+5 \\
& =7
\end{aligned}\)

4. f(x) = 6 + 10x – 7x²

Solution:

Given

f(x) = 6 + 10x – 7x²

\(\begin{aligned}
therefore f(1) & =6+10.1-7(1)^2 \\
& =6+10-7 \\
& =9 \\
f(-1) & =6+10(-1)-7(-1)^2 \\
& =6-10-7 \\
& =-11
\end{aligned}\)

Class 9 Maths WBBSE

Question 3. Let us check the following statements –

1. The zero of the polynomial P(x) = x – 1.

Solution: P(x)=x-1=0
∴ X = 1

2. The zero of the polynomial P(x) = 3-x is 3.

Solution: P(x)=3-x=0
∴ -x=-3 or, x = 3
∴ X=3,

3. The zero of the polynomial P(x) = 5x + 1 is

Solution: P(x)=5x+1=0
∴ 5x + 1 = 0 or, 5x = -1
∴ x = -1/5

4. The two zeroes of the polynomial P(x) = x² -9 are 3 and -3.

Solution: P(x) = x² -9=0
∴ x2-9=0 or, x2 = 9
or, x = ±3
∴ The two zeroes of the polynomial P(x) are 3 and -3.

5. The two zeroes of the polynomial P(x) = x²-5x are 0 and 5. Solve: P(x) = x²-5x=0

Solution: P(x) = x²-5x=0
∴ x² – 5x = 0 or, x(x-5)=0
or, x = 0 and, x-5=0; x = 5
∴ x = 0, 5
∴The two zeroes of the polynomial P(x) are 0 and (5).

6. The two zeroes of the polynomial P(x) = x²-2x-8 are 4 and (-2).

Solution: P(x) = x²-2x-8=0
∴ x²-2x-8=0
or, x²-4x+2x-8=0
or, x(x-4)+2(x-4)=0
or, (x-4) (x+2)=0
x = 0 and x + 2 = 0
∴ x = 4 and x = -2
∴ P(x) is a polynomial whose two zeroes are 4 and (-2).

Class 9 Maths WBBSE

Question 4. Let us determine the zeroes of the following polynomials –

1. f(x) = 2-x

Solution:

Given

f(x)=2-x … f(x) = 0
∴ 2-x=0
or, -x=-2
or, x = 2
∴ f(x) is a polynomial whose zero is 2.

2. f(x) = 7x + 2

Solution:

Given

f(x)=7x+2
∴ f(x) = 0
∴7x+2=0
or, 7x=-2
or, x = -2/7
∴ f(x) is a polynomial whose zero is -2/7

3. f(x) = x + 9

Solution:

Given

f(x)=x+9
∴f(x) = 0
∴x+9=0
or, x=-9
∴ f(x) is a polynomial whose zero is (-9).

4. f(x) = 62x

Solution:

Given

f(x) = 6-2x
∴ f(x) = 0
∴6-2x=0
or, -2x=-6
or, 2x = 6
or, x = 3
∴ f(x) is a polynomial whose zero is 3.

5. f(x) = 2x

Solution:

Given

f(x) = 2x
∴ f(x) = 0
∴2x = 0
or, x = 0
∴ f(x) is a polynomial whose zero is 0.

6. f(x) = ax + b, (a = 0)

Solution:

Given

f(x) = 0
∴ax + b = 0
∴ X = -a/b
∴ f(x) is a polynomial whose zero is (-b/a)

Class 9 Mathematics West Bengal Board Chapter 7 Polynomial Exercise 7.3

1. By applying Remainder Theorem, let us calculate and write the remainder that I shall get in each case, when x³-3x²+ 2x + 5 is divided by:

1. x-2

Solution: x-2
The zero of the linear polynomial x 2 = 0
∴ X = 2
From the Remainder Theorem, division of f(x) = x³-3x²+ 2x + 5 by (x-2) gives the remainder f(2).

∴ The required remainder = f(2) =(2)3-3 (2)² +2.2 +5 = 8-12+ 4+ 5 = 17-12 = 5

2. x+2

Solution: The zero of the linear polynomial x + 2 = 0
∴ X=-2
The required remainder = f(-2) =(-2)³-3(-2)² + 2 (-2)+5 =-8-12-4+5 =-19

3. 2x-1

Answer: The zero of the linear polynomial 2x-1=0 ∴x =1/2

\(\begin{aligned}
& =f\left(\frac{1}{2}\right) \\
& =\left(\frac{1}{2}\right)^3-3\left(\frac{1}{2}\right)^2+2 \cdot \frac{1}{2}+5 \\
& =\frac{1}{8}-\frac{3}{4}+1+5
\end{aligned}\)

 

\(\begin{aligned}
& =\frac{1}{8}-\frac{3}{4}+6 \\
& =\frac{1-6+48}{8} \\
& =\frac{43}{8} \\
& =5 \frac{3}{8}
\end{aligned}\)

 

4. 2x+1

Solution:

\(\begin{aligned}
& =f\left(-\frac{1}{2}\right) \\
& =\left(-\frac{1}{2}\right)^3-3\left(-\frac{1}{2}\right)^2+2\left(-\frac{1}{2}\right)+5 \\
& =-\frac{1}{8}-\frac{3}{4}-1+5 \\
& =\frac{-1-6-8+40}{8} \\
& =\frac{40-15}{8}=\frac{25}{8}=3 \frac{1}{8}
\end{aligned}\)

 

Question 2. By applying Remainder Theorem, let us calculate and write the remainders that I shall get when the following polynomials are divided by (x-1).

1. x³– 6x² + 13x + 60

Solution: Zero of the linear polynomial x -1 = 0
∴ X = 1
= f(1) = (1)³-6(1)² + 13.1 +60
= 1-6+13+60 =74-6=68

2. x³-3x² + 4x + 50

Solution: Let f(x) = x³-3x² + 4x + 50
= f(1) =(1)³-3(1)²+4.1 +50 = 1-3+4+ 50= 55 – 3 = 52

(3)4x³ + 4x²-x-1

Solution: Let f(x) = 4x³ + 4x²-x-1
= f(1)= 4(1)³-4(1)²-1-1= 4+4-1-1=8-2. = 6

(4)11x³– 12x² – x + 7

Solution: Let f(x) =11x³– 12x² – x + 7
= f(1) =11(1)³ – 12(1)²-1+7 11121+7 = 18 – 13= 5

Question 3. Applying Remainder Theorem, let us write the remainders, when –

1. The polynomial (x³-6x² + 9x-8) is divided by (x-3)

Solution: x-3=0
∴ X=3
(x) = x³-6x² + 9x-8

∴ Remainder = f(3) = (3)³-6(3)²+9(3)-8 =2754 +27-8 = 54-62=-8

2. The polynomial (x³ -ax² + 2x -a) is divided by (x-a).

Solution: x-a=0
∴ X = a
f(x) = x³ -ax² + 2x -a

∴ Remainder = f(a) = (a)³– a(a)² + 2.a – a = a³-a³+2a-a=a

Question 4. Applying Remainder Theorem, let us calculate whether the polynomial p(x) = 4x³+4x²-x -1 is a multiple of (2x + 1) or not.

Solution: 2x+1=0
∴ X=-1/2

∴ The zero of the linear polynomial (2x + 1) = -1/2
p(x) = 4x³+4x²-x -1
(2x+1) is a factor of P(x) if P(-1/2)=0

\(\begin{aligned}
& P\left(-\frac{1}{2}\right)=4\left(-\frac{1}{2}\right)^3+4\left(-\frac{1}{2}\right)^2-\left(-\frac{1}{2}\right)-1 \\
& =4\left(-\frac{1}{8}\right)+4\left(\frac{1}{4}\right)+\frac{1}{2}-1 \\
& =\frac{-1}{2}+1+\frac{1}{2}-1 \\
& =0
\end{aligned}\)

 

Question 5. For what value of a, the divisions of two polynomials (ax³+3x²-3) and (2x³-5x+a) by (x-4) give the same remainder – let us calculate and write it.

Solution: Let f(x) = ax³+3x²-3 and g(x) = 2x³– 5x + a.
If f(x) is divided by (x-4) the remainder is

\(\begin{aligned}
& f(4)=a(4)^3+3(4)^2-3 \\
& =64 a+48-3 \\
& =64 a+45
\end{aligned}\)

If g(x) is divided by'(x-4) the remainder is

\(\begin{aligned}
g(4) & =2(4)^3-5.4+a \\
& =128-20+a \\
& =108+a
\end{aligned}\)

∴ f(4) = g(4)
∴ 64a+45=108+a
or, 64a-a=108-45
or, 63a= 63
or, a = 1.
∴The value of a = 1

Question 6. The two polynomials x³ + 2x²-px-7 and x³ + px² – 12x + 6 are divided by (x + 1) and (x-2) respectively and if the remainder R₁ and R₂ are obtained such that 2R₁ + R₂=6, then let us calculate the value of p.

Solution:

Given

The two polynomials x³ + 2x²-px-7 and x³ + px² – 12x + 6 are divided by (x + 1) and (x-2) respectively and if the remainder R₁ and R₂ are obtained such that 2R₁ + R₂=6,

Let f(x) =x³ + 2x²-px-7
If f(x) is divided by (x + 1) the remainder is f(-1)=(-1)³ + 2(-1)²-P(-1)-7 =-1+2+P-7 =P-6
According to 1st condition, R₁ = P-6 …..(1)

Again, let g(x) = x³ + px² – 12x + 6
If g(x) is divided by (x-2) the remainder is g(2) = (2)³ + P(2)² – 12.2 +6 =8+4P-24 + 6 = 4P-10
According to 2nd condition, R₂ = 4P – 10 ………(2)

∴ 2R₁ + R₂ = 6
or, 2(P-6)+4P-10=6 or, 2P-12+ 4P-10=6
or, 6P 22=6
or, 6P=6+22
or, 6P=28

\(or, P=\frac{28}{6} or, P=\frac{14}{3}=4 \frac{2}{3}\)

∴The value of \(P=4 \frac{2}{3}\)

Question 7. The polynomial x⁴ – 2x³ + 3x²– ax + b is divided by (x-1) and (x + 1) and the remainders are 5 and 19 respectively. But if that polynomial is divided by x + 2, then what will be the remainder – Let us calculate.

Solution:

Given

The polynomial x⁴ – 2x³ + 3x²– ax + b is divided by (x-1) and (x + 1) and the remainders are 5 and 19 respectively. But if that polynomial is divided by x + 2,

Let f(x) = x⁴ – 2x³ + 3x²– ax + b
If f(x) is divided by (x-1) the remainder is f(1) =(1)⁴-2(1)³+3(1)-a.1+b = 1-2+3- a+b =2-a+b

According to 1st condition, 2-a+b=5
or,a+b=5-2
or, a – b = -3…(1)

Again, if f(x) is divided by (x + 1) the remainder is
f(-1)=(-1)⁴-2(-1)³+3(-1)²-a(-1)+b =1+2+3+a+b = a+b+6

According to 2nd condition, a+b+6=19
or, a+b=19-6
or, a+b=13 ……. (2)

from the eqation (2)/(1)
a+b=13/a – b = -3 = 2a =10

Adding, 2a = 10
or, a =10/2
or,a = 5

Putting value 4 in equation (2) we get, 5+b=13
b=13-5=8
∴ f(x) = x⁴ – 2x³ + 3x²– 5x + 8

If f(x) is divided by (x + 2) the remainder is
f(-2)=(-2)⁴ -2 (-2)³ + 3(-2)² -5(-2)+8 = =16+16 +12 + 10 + 8 = 62
The required remainder = 62.

Question 8. If \(f(x)=\frac{a(x-b)}{a-b}+\frac{b(x-a)}{b-a}\) then let us show that f(a) + f(b) = f(a + b).

Solution:

Given

\(f(x)=\frac{a(x-b)}{a-b}+\frac{b(x-a)}{b-a}\) \(\begin{aligned}
& therefore f(a)=\frac{a(a-b)}{a-b}+\frac{b(a-a)}{b-a} \\
& =a+0 \\
& =a \\
& f(b)=\frac{a(b-b)}{a-b}+\frac{b(b-a)}{b-a} \\
& =0+b \\
& =b \\
&∴f(a)+f(b)=a+b \\
&
\end{aligned}\)

 

Again, \(\begin{aligned}
f(a+b) & =\frac{a(a+b-b)}{a-b}+\frac{b(a+b-a)}{b-a} \\
& =\frac{a^2}{a-b}+\frac{b^2}{-(a-b)} \\
& =\frac{a^2}{a-b}-\frac{b^2}{a-b} \\
& =\frac{a^2-b^2}{a-b} \\
& =\frac{(a+b)(a-b)}{(a-b)} \\
& =a+b
\end{aligned}\)

Question 9. If f(x) = ax + b and f(0) = 3, f(2) = 5, then let us determine the values of a and b.

Solution:

Given

If f(x) = ax + b and f(0) = 3, f(2) = 5

f(x) = ax + b
∴f(0) = a.0+ b = 0 + b = b
∴b = 3
Again, f(2) = a.2 + b = 2a + b
∴2a + b = 5
or, 2a+35 [∴ b = 3]
or, 2a=5-3
or, 2a = 2
or, a = 1
∴ a = 1 and b = 3

Question 10. If f(x) = ax² + bx + c and f(0) = 2, f(1) = 1, and f(4) = 6, then let us calculate the values of a, b, and c.

Solution:

Given

f(x) = ax² + bx + c
∴ f(0) = a(0)²+ b.0+ c =0+0+c=c
∴ c = 2 ….(1)

Again, f(1)= a(1)²+b.1+ c= a+b+c
∴ a+b+c=1
or, a+b+2=1(∴c=2)
or, a+b=1-2
or, a+b=-1 ……(2)

Again, f(4) = a(4)2 + b.4 + c = 16a+ 4b + C
16a+ 4b+c=6
or, 16a+ 4b+2=6(c=2)
or, 16a+ 4b6-2
or, 16a+ 4b4
or, 4(4a + b) = 4

or, \(4 a+b=\frac{4}{4}\)

4a + b = 1 …..(3)

Divide equation (3)/(2)

\(\begin{aligned}
& 4 a+b=1 \\
& a+b=-1 \\
& (-) \quad(-) \quad(+) \\
& 3 a \quad=2
\end{aligned}\)

or, a =2/3

Putting the value of a in equation (2),

\(\frac{2}{3}+b=-1\)

or, \(b=-1-\frac{2}{3}\)

or,\(b=-\left(\frac{3+2}{3}\right)\)

or,\(b=\frac{-5}{3}\)

∴\(a=\frac{2}{3}, b=\frac{-5}{3}, c=2\)

Question 11. Multiple Choice Questions

1. Which of the following is a polynomial in one variable?

1. \(x+\frac{2}{x}+3\)

2. \(3 \sqrt{x}+\frac{2}{\sqrt{x}}+5\)

3. \(\sqrt{2} x^2-\sqrt{3} x+6\)

4. \(x^{10}+y^5+8\)

Solution: 3. \(\sqrt{2} x^2-\sqrt{3} x+6\)

2. Which of the following is a polynomial?

1. x-1

2. \(\frac{x-1}{x+1}\)

3. \(x^2-\frac{2}{x^2}+5\)

4. \(x^2+\frac{2 x^{\frac{3}{2}}}{\sqrt{x^2}}+6\)

Solution: 1. x-1

3. Which of the following is a linear polynomial?

1. x + x²
2. x + 1
3. 5x²-x+3
4. \(x+\frac{1}{x}\)

Solution:2. x + 1

4. Which of the following is a second-degree polynomial?

1. √x-4
2. x³ + x
3. x³+ 2x + 6
4. x²+ 5x + 6

Solution: 4. x²+ 5x + 6

5. The degree of the polynomial √3 is

1. 1/2
2.  2
3. 1
4. 0

Solution: 4. 0

Question 12. Short answer type questions :

1. Let us write the zero of the polynomial p(x) = 2x – 3.

Solution: 2x-3=0
∴ X =3/2
∴The zero of polynomial p(x) = 2x – 3 is 3/2.

2. If p(x) = x + 4, let us write the value of p(x) + P(-x).

Solution: p(x) = x+4
∴ p(x) + p(x) = x+4x+4=8.

3. Let us write the remainder, if the polynomial x³ + 4x² + 4x – 3 is divided by x.

Solution: Let f(x) =x³ + 4x² + 4x – 3
∴ The required remainder =f(0) = (0)³ + 4(0)²+4.0-3=-3.

4. If \((3 x-1)^7=a_7 x^7+a_6 x^6+a_5 x^5+\ldots \ldots \ldots+a_1 x+a_0\) then let us write the value of \(a_7+a_6+a_5+\ldots \ldots \ldots+a_0\) (where a₇, a₆……….a₀ are constants).

Solution: \((3 x-1)^7=a_7 x^7+a_6 x^6+a_5 x^5+\ldots \ldots \ldots+a_1 x+a_0\)

or, (3.1-1)⁷ = a₇(1)7+ a₆(1)⁶ + a₅(1)⁵+. . ..+a₁x+a₀ (Putting x = 1 on both sides)
∴ (2)⁷ = a₇+ a₆+ a₅… a₁+ a₀
∴ a₇+ a₆ + a₅ + ……. a₁+ a₀ = 128.

 

Chapter 7 Polynomial Exercise 7.4

 

Question 1. Let us calculate and write, which of the following polynomials will have a factor (x+1).

Solution: x + 10 ∴x=-1
∴ The zero of polynomial (x + 1) is – 1.

1. 2x³ + 3x² – 1

Solution: Let f(x) =2x³ + 3x² – 1
∴f(-1) =2(-1)³+3(-1)²-1 =-2+3-1=3-3 = 0
∴ The factor of 2x³ + 3x² – 1 is (x + 1).

2. x⁴ + x³– x² + 4x + 5

Solution: Let f(x) = x⁴ + x³– x² + 4x + 5
∴ f(-1)=(-1)⁴+(-1)³-(-1)² + 4(-1)+5 1-1-1-4+5 =6-6 = 0
∴ One factor of x⁴ + x³– x² + 4x + 5 is x + 1.

3. 7x³ + x² + 7x + 1

Solution: Let f(x) = 7x³ + x² + 7x + 1 =7(-1)³+(-1)²+7(-1)+1 =-7+1-7+1 = 2 – 14 = 12
∴ (x + 1) is a factor of (7x³ + x² + 7x + 1).

4. 3+3x-5x³– 5x⁴

Solution: Let f(x)=3+3x-5x³– 5x⁴
f(-1)= 3+3(-1)-5(-1)³-5(-1)⁴ = 3-3+5-5 = 8-8 = 0
∴ One factor of 3+3x-5x³– 5x⁴ is (x + 1).

5. x⁴ + x² + x +1

Solution: Let f(x) =x⁴ + x² + x +1
f(-1) = (-1)⁴+(-1)²+(-1)+1 =1+1 1+1 = 3-1 = 2
∴ (x + 1) is a factor of (x⁴ + x² + x +1).

6. x³ + X² + X +1

Solution: f(-1) = (-1)³+(-1)²+(-1)+1=-1+1+1+1=2-2 = 0
∴ One factor of x³ + X² + X +1 is (x + 1).

Question 2. By using Factor Theorem, let us write whether g(x) is a factor of the following polynomials f(x):

1. f(x) = x⁴-x²-12 and g(x) = x + 2

Solution: g(x) = x + 2 is a factor of f(x) if f(-2) = 0
∴ f(x)= x⁴-x²-12 = (-2)⁴-(-2)²-12=16- 4- 12 = 16- 16 = 0
∴ g(x) is a factor of f(x).

2. f(x) = 2x³ +9x²-11x-30 and g(x) = x + 5

Solution: x+5=0
∴ X=-5
∴ f(x) = 2x³ +9x²-11x-30
f(-5) = 2(-5)³ +9(-5)²-11x-30=250+225 +55 – 30 = 280 280= 0
∴ g(x) is a factor of f(x).

3. f(x) = 2x³ + 7x²-24x-45 and g(x) = x – 3

Solution: x-3=0
∴ X=3
∴ f(x) = 2x³ + 7x²-24x-45
∴ f(3) = 2(3)³+7(3)²-24.3-45=54+63 72-45= 117 117 = 0
∴ g(x) is a factor of f(x).

4. f(x) = 3x³ + x² – 20x + 12 and g(x) = 3x – 2

Solution:3x-2=0
∴ f(x) = 3x³ + x² – 20x + 12

\(\begin{aligned}
f\left(\frac{2}{3}\right) & =3\left(\frac{2}{3}\right)^3+\left(\frac{2}{3}\right)^2-20 \cdot \frac{2}{3}+12 \\
& =3 \cdot \frac{8}{27}+\frac{4}{9}-\frac{40}{3}+12 \\
& =\frac{8}{9}+\frac{4}{9}-\frac{40}{3}+12 \\
& =\frac{8+4-120+108}{9} \\
& =\frac{120-120}{9}=\frac{0}{9}=0
\end{aligned}\)

∴ g(x) is a factor of f(x).

Question 3. Let us calculate and write the value of k for which the polynomial 2x + 3×3+2kx2 + 3x + 6 is divided by x + 2.
Solution: Let f(x) = 2x + 3x³+2kx² + 3x + 6
The zero of (x + 2) is – 2.
∴ (x+2) is a factor of f(x).
∴ f(-2)=0
∴ 2(-2)⁴+3(-2)³+2.k.(-2)²+3(-2)+6=0
or, 32-24+8k-6+6=0
or, 8+ 8k = 0
or, 8k=-8
or, k=-1

Question 4. Let us calculate the value of k for which g(x) will be a factor of the following polynomials f(x):

1.  f(x) = 2x³ +9x²+ x + k and g(x) = x-1

Solution: g(x) = (x-1) is a factor of f(x) if x = 1 (x-1=0. x = 1)
∴ g(x) is a factor of f(x)
∴ f(1) = 0
∴ 2(1)³+9(1)²+1+k=0
or, 2+9+1+k=0
or, k=-12
∴ If k = 12, g(x) is a factor of f(x).

2. f(x)=kx²-3x+k and g(x)=x-1

Solution: g(x) = (x-1) is a factor of f(x) if x = 1 (x-1=0. x = 1)
∴g(x) is a factor of f(x)
∴f(1) = 0
∴ k(1)²-3.1+k=0
or, 2k-3=0
or, k = 3/2
∴ If k = 3/2,g(x) is a factor of f(x).

3. f(x) =2x⁴ + x³ – kx²-x+6 and g(x) = 2x-3

Solution: g(x) = 2x-3.
∴ The zero of g(x) is 3/2(2x-3=0 ∴X =3/2)
∴ g(x) is a factor of f(x)

\(\begin{aligned}
& f\left(\frac{3}{2}\right)=0 \\
&2\left(\frac{3}{2}\right)^4+\left(\frac{3}{2}\right)^3-k\left(\frac{3}{2}\right)^2-\frac{3}{2}+6=0 \\
& \text { or, } 2 \cdot \frac{81}{16}+\frac{27}{8}-\frac{9 k}{4}-\frac{3}{2}+6=0 \\
& \text { or, } \frac{81}{8}+\frac{27}{8}-\frac{9 k}{4}-\frac{3}{2}+\frac{6}{1}=0 \\
& \text { or, } \frac{81+27-18 k-12+48}{8}=0 \\
& \text { or, } 144-18 k=0 \\
& \text { or, }-18 k=-144 \\
& \text { or, } k=\frac{-144}{-18} \\
& k=8
\end{aligned}\)

∴If k= 8 then g(x) will be a footer of W

4. f(x) =2x³ + kx² + 11x+ k + 3 and g(x) = 2x – 1

Solution: The zero of polynomial g(x) = (2x-1) is 1/2(2x-1=0 ∴x = 1/2)
∴ g(x) is the factor of f(x)

\(\begin{aligned}
& f\left(\frac{1}{2}\right)=0 \\
& 2\left(\frac{1}{2}\right)^3+K\left(\frac{1}{2}\right)^2+11 \cdot \frac{1}{2}+K+3=0
\end{aligned}\)

 

\(\text { 2. } \frac{1}{8}+k \cdot \frac{1}{4}+\frac{11}{2}+k+3=0\)

 

\(\frac{1}{4}+\frac{k}{4}+\frac{11}{2}+k+3=0\)

 

\(\frac{1+k+22+4 k+12}{4}=0\)

 

or, 5k + 35 = 0
or, 5k = -35
or, k = -35/5 = -7
∴ If k=-7, g(x) is a factor of f(x).

Question 5. Let us calculate and write the values of a and b if x²-4 is a factor of the polynomial ax⁴ + 2x³-3x²+ bx-4.

Solution: x²-4=0
or, (x+2) (x-2)=0
x+2=0 or, x-2=0
Let f(x) = ax⁴ + 2x³-3x²+ bx-4

∴ (x²-4) is a factor of f(x)
∴ f(2) = 0 and f(-2) = 0
f(2) = 0
∴ a(2)⁴+2(2)³-3(2)²+ b.2-4=0
or, 16a+ 16-12+2b-4=0
or, 16a+2b=0
or, 8a+ b = 0 …..(1)

f(-2)=0
∴a(-2)⁴+2(-2)³-3(-2)²+ b.(-2)-4=0
or, 16a-16-12-2b-4=0
or, 16a-2b= 32
or,8a-b=16 ….(2)

8a+ b = 0

Divide by the equation (2)/(1)

\(\begin{aligned}
& 8 a-b=16 \\
& 8 a+b=0
\end{aligned}\)

 

or, a =16/16
or, a =1

Putting the value of a in equation (1),
8.1+b=0
or, b=-8
∴ a = 1, b=-8

Question 6. If (x + 1) and (x+2) are two factors of the polynomial x³ + 3x²+2ax + b, then let us calculate and write the values of a and b.

Solution:

Given

If (x + 1) and (x+2) are two factors of the polynomial x³ + 3x²+2ax + b

Let f(x) = x³ + 3x²+2ax + b
The zero of (x + 1) is -1 (x+1=0  ∴x=-1)
∵ (x + 1) is a factor of f(x)
∴ f(-1)-0
∴ (-1)³+3(-1)²+2a(-1)+b=0
or, -1+3-2a+b=0
or, 2-2a+b=0
or, 2a+b=-2
or, 2a-b=2 ….(1)

Again, the zero of (x + 2) is -2
∵ (x+2) is a factor of f(x)
∴ f(-2)=0
∴ (-2)³+3(-2)²+2a(-2)+b=0
or, -8+12-4a+b=0
or, 4-4a+b=0
or, 4a+b=-4
or, 4a-b = 4 …(2)

(2)-(1)

\(\begin{aligned}
& 4 a-b=4 \\
& 2 a-b=2 \\
& (-)(+)(-) \\
& 2 a=2
\end{aligned}\)

Subtracting, 2a=2
or, a = 1

Putting the value of in equation (1), 2.1-b=2
or, -b = 2-2
or, – b = 0
or, b = 0
∴ a = 1, b = 0.

Question 7. If the polynomial ax³ + bx²+x-6 is divided by x-2 and the remainder is 4, then et us calculate the values of a and b when x + 2 is a factor of this polynomial.

Solution:

Given

If the polynomial ax³ + bx²+x-6 is divided by x-2 and the remainder is 4

Let f(x) = ax³ + bx²+x-6
∴ If f(x) is divided by (x-2), the remainder is 4
∴f(2) = 4[ The zero of (x-2) is 2]
or, 8a+ 4b-4 = 4
or, 4(2a + b) = 4+4
or, (2a + b) = 8/4
or, 2a + b = 2

Again, (x+2) is a factor of f(x)
∴ f(-2) = 0 (∵x+2=0 ∴x=-2)
∴ a(-2)³+(-2)2+(-2)-6-0
or, -8a+ 4b-2-6=0
or, -8a+ 4b = 8
or, -4(2a – b) = 8

or, \(2 a-b=\frac{8}{-4}\)

or, 24-b =-2 ….(2)

Adding, (2)+(1)

\(\begin{aligned}
& 2 a-b=-2 \\
& 2 a+b=2 \\
& 4 a=0
\end{aligned}\)

or, a = 0
∴ 2.0 + b = 2
or, 0+b=2
or, b = 2
∴ a = 0, b = 2

Question 8. Let us show that if n is any positive integer (even or odd), x-y is a factor of the polynomial xⁿ – yⁿ

Solution: x – y = 0
∴ x = y
Let f(x) = xⁿ – yⁿ
∴f(y) = yⁿ – yⁿ= 0
(x-y) is a factor of xⁿ – yⁿ.

Question 9. Let us show that if n is any positive odd integer, then x + y is a factor of xⁿ + yⁿ.

Solution: x + y = 0
∴ x = – Y
Let f(x) = xⁿ + yⁿ
∴ f(-y)= (-y)ⁿ+ yⁿ =-yⁿ+ yⁿ= 0
∴ (x + y) is a factor of xⁿ + yⁿ.

Question 10. Let us show that if n be any positive integer (even or odd), then x – y will never be a factor of the polynomial xⁿ + yⁿ

Solution: x-y=0
∴ x = y
Let f(x) = xⁿ + yⁿ
f(y) = (y)ⁿ+ yⁿ
f(y) = 2yⁿ ≠0
∴ (x-y) can never be a factor of f(x) = (xⁿ + yⁿ).

Question 11. Multiple Choice Questions

1. x³ + 6x² + 4x + k (x + 2)

(1)-6
(2)-7
(3)-8
(4) 10

Solution: Let f(x) = x³ + 6x² + 4x + k
The zero of (x+2) is -2
∴ (x+2) is a factor of f(x)
∴ f(-2) = 0
∴ (-2)³+6(-2)²+4(-2)+k=0
or, 8+24-8+k=0
or, 8+ k = 0.
or, k=-8

∴(3) – 8

2. In the polynomial f(x) if \(f\left(-\frac{1}{2}\right)=0\)  then a factor of f(x) will be

(1) 2x-1
(2) 2x + 1
(3) x-1
(4) x + 1

Solution: \(f\left(-\frac{1}{2}\right)=0\)

∴X =-1/2
or, 2x = -1
or, 2x+1=0
∴(2x+1) is a factor of f(x).

∴ (2)2x + 1

3. (x-1) is a factor of the polynomial f(x) but it is not a factor of g(x). So (x-1) will be a factor of

(1)f(x) g(x)
(2)- f(x) + g(x)
(3)f(x) = g(x)
(4){f(x) + g(x)}g(x)

Solution: (x-1) is a factor of f(x) g(x)
∴ (1)f(x) g(x)

4. (x + 1) is a factor of the polynomial xⁿ + 1 when

(1) n is a positive odd integer
(2) n is a positive even integer
(3) n is a negative integer
(4) n is a positive integer

Solution: We know if n is an odd positive integer, (x+y) is a factor of (xⁿ + yⁿ )
∴ (x + 1) is a factor of xⁿ+ 1, i.e., xⁿ+1ⁿ when n is an odd positive integer.

∴(1) n is a positive odd integer

5. If n²-1 is a factor of the polynomial an⁴ + bn³ + cn² + dn + e

(1) a +c+e=b+d
(2) a+b+e=c+d
(3) a+b+c=d+e
(4) b+c+d=a+e

Solution: Let f(n) = an⁴ + bn³ + cn² + dn + e
Factor of 1(n) = n²-1-(n + 1) (n-1)
∴ f(-1)=0 and f(1) = 0 When f(-1) = 0.

or, a(-1)⁴+b(-1)³+c(-1)² +d(-1)+e=0
or, a-b+c-d+e=0
∴ a+c+e=b+d

Again, f(1)=0
∴ a(1)⁴+b(1)³+c(1)²+d.1+e=0
or, a+b+c+d+e=0
∴ a+c+e-b-d

∴(1) a +c+e=b+d

Question 12. Short answer type questions:

1. Let us calculate and write the value of a for which x + a will be a factor of the polynomial x³+ax²-2x+a-12.

Solution: Let f(x) = x³+ax²-2x+a-12
Factor of f(x) is (x + a)
∴f(-a) = 0
∴(-a)³ + a(-a)²-2(-a) + a-12=0
or,(-a)³+ a³+2a + a-12=0
or, 3a= 12
or, a = 12/3
or, a = 4

2. Let us calculate and write the value of k for which x-3 will be a factor of the polynomial k²x³-kx²+3kx-k.

Solution: Let f(x) = k²x³-kx²+3kx-k
Factor of f(x) is (x-3)
∴f(3) = 0
∴k²(3)³-k(3)²+3k.3-k=0
or, 27k²-9k+9k – k=0
or, 27k²-k=0
or, k(27k-1)=0 = k = 0 and 27k10.
∴ k = 1/27
∴ k = 0, 1/27

3. Let us write the value of f(x) + f(-x) when f(x) = 2x + 5.

Solution: f(x) = 2x + 5
∴ f(x) + f(x) = 2x+5+2(-x)+5 =2x+5-2x+5=10

4. Both (x-2) and \(\left(x-\frac{1}{2}\right)\) are factors of the polynomial px² + 5x + r, let us calculate and write the relation between p and r.

Solution: Let f(x) = px² + 5x + r
Factor of f(x) is (x-2)
∴ f(2) = 0 (x-2=0; x=2) .
∴ p(2)²+ 5.2 + r=0
or, 4p+r+10=0
or, 4p+r=-10  ….(1)

Again, factor of f(x) is (x) is \(\left(x-\frac{1}{2}\right)\)

\(\begin{aligned}
&  f\left(\frac{1}{2}\right)=0\left(because x-\frac{1}{2}=0 therefore x=\frac{1}{2}\right) \\
& p\left(\frac{1}{2}\right)^2+5 \cdot \frac{1}{2}+r=0 \\
& \text { or, } \frac{p}{4}+\frac{5}{2}+r=0 \\
& \text { or, } \frac{p+10+4 r}{4}=0
\end{aligned}\)

or, p +4r + 10 = 0

or, p + 4r = -10 .,…..(2)

(1) and (2)

4p + r = p + 4r
or, 4p – p = 4r -r
or, 3p = 3r
or, p = r

5. Let us write the roots of the linear polynomial f(x) = 2x + 3. Solve: 2x+3=0

Solution: 2x=-3
or, 2x = -3
or, x = -3/2
∴ The roots of f(x) = 2x + 3 = -3/2