AP 10th Class Maths Bits Chapter 3 Polynomials with Answers

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AP SSC 10th Class Maths Bits 3rd Lesson Polynomials with Answers

Question 1.
Write the sum of zeroes of
bx² + ax + c.
Answer:
\(\frac{-a}{b}\)

Question 2.
Find product of the zeroes of
p(x) = (x-2).(x + 3)
Answer:
= -6
Explanation:
p(x) = x² + x – 6,
Product of zeroes = \(\frac{\mathrm{c}}{\mathrm{a}}\) = – 6.

Question 3.
5x – 3 represents which type of poly¬nomial ?
Answer:
Linear.

Question 4.
Find the value of ‘x’ which satisfies 2(x – 1) – (1 – x) = 2x + 3 ?
Answer:
6
Explanation:
2x – 2 – 1 + x = 2x + 3
⇒ 3x – 2x = 3 + 3 = 6
⇒ x = 6.

Question 5.
Write the degree of the polynomial
\(\sqrt{2}\)x² – 3x + 1.
Answer:
2

Question 6.
Write order of the polynomial 5x⁷ – 6x⁵ + 7x – 6.
Answer:
7

Question 7.
Find the product of zeroes of 2x² – 3x + 6.
Answer:
3
Explanation:
Product of zeroes = \(\frac{c}{a}=\frac{6}{2}\) = 3.

Question 8.
Find sum of zeroes of a polynomial x³ – 2x² + 3x – 4.
Answer:
2
Explanation:
Sum of zeroes = α + β + γ
= \(\frac{-b}{a}=\frac{-(-2)}{1}\) = 2

Question 9.
Find a quadratic polynomial, the sum of whose zeroes is zero and one zero is 4, is
Answer:
x² – 16 = 0
Explanation:
Sum of zeroes = \(\frac{-b}{a}\) = 0,
α + β = 0, β = 4
⇒ α = -4 x2 – (α + β)x + αβ = 0
⇒ x² – 0(x) -16 = 0
⇒ x² – 16 = 0.

Question 10.
If p(x) = x² – ax – 3 and p(2) = – 3, then find Answer:
Answer:
2
Explanation:
p(x) = (2)² – 2a – 3 = – 3
⇒ 1 — 2a = — 3
⇒ 4 – 2a = 0
⇒ 4 = 2a ⇒ a = 2

Question 11.
Write the zero value of polynomial px + q.
Answer:
\(\frac{-\mathrm{q}}{\mathrm{p}}\)

Question 12.
In a division, if divisor is x + 1, quo¬tient is x and remainder is 4, then find dividend.
Answer:
x (x + 1) + 4 = x² + x + 4
Explanation:
Dividend = Divisor x Quotient
+ Remainder
= (x + 1) x + 4 = x² + x + 4.

Question 13.
Find the zero value of linear polynomial ax – b.
Answer:
\(\frac{\mathrm{b}}{\mathrm{a}}\)

Question 14.
Find the sum of the zeroes of the poly-nomial x² + 5x + 6.
Answer:
– 5

Question 15.
4y² – 5y + 1 is an example for this type of polynomial.
Answer:
Quadratic polynomial.

Question 16.
Write the degree of the polynomial 5x⁷ – 6x⁵ + 7x – 4.
Answer:
7

Question 17.
How much the sum of zeroes of the polynomial 2x² – 8x + 6 ?
Answer:
4

Question 18.
f α,β are the zeroes of x² + x +1, then find \(\frac{1}{\alpha}+\frac{1}{\beta}\)
Answer:
-1
Explanation:
α + β = -1, αβ = 1
⇒ \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{-1}{1}\) = -1

Question 19.
Write the number of zeroes of the poly¬nomial in the given graph.

Answer:
3

Question 20.
Write product of zeroes of the cubic polynomial 3x³ – 5x² – 11x – 3.
Answer:
1
Explanation:
Product of zeroes = αβγ
= \(\frac{-d}{a}=\frac{-(-3)}{3}=\frac{3}{3}\) = 1

Question 21.
The following is the graph of a poly¬nomial. Find the zeroes of the poly¬nomial from the given graph.

Answer:
– 2,1

Question 22.
Find the value of p(x) = 4x² + 3x + 1 at x = -1.
Answer:
2

Question 23.
If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d and (a ≠ 0), then find αβγ

Question 24.
4x + 6y = 18 doesn’t pass through ori-gin, then its graph indicates
Answer:
A straight line

Question 25.
If ‘3’ is one of the zeroes of
p(x) = x² + kx – 9, then find the value of k.
Answer:
0
Explanation:
p(3) = 3² + 3k – 9 = 0
⇒ 3k = 0 ⇒ k = 0

Question 26.
When p(x) = x² – 8x + k leaves a re-mainder when it is divided by (x – 1), then find k.
Answer:
k > 7
Explanation:
p(1) = 1 – 8 + k = 0
⇒ k > 7

Question 27.
If α, β, γ are zeroes of x³ + 3x² – x + 2, then find the value of αβγ.
Answer:
-2

Question 28.
Make a quadratic polynomial having 2, 3 as zeroes.
Answer:
x² – 5x + 6 = 0
Explanation:
x² – (2 + 3)x + 2 . 3 = 0
⇒ x² – 5x + 6 = 0.

Question 29.
Write a quadratic polynomial, whose zeroes are \(\sqrt{2}\) and –\(\sqrt{2}\).
Answer:
x² – 2 = 0
Explanation:
x² – \((\sqrt{2}-\sqrt{2}) x+\sqrt{2}(-\sqrt{2})\) = 0
⇒ x² – 0(x) -2 = 0
⇒ x² – 2 = 0.

Question 30.
Find the coefficient of x7 in the poly¬nomial 7x¹⁷ – 17x¹¹ + 27x⁵ – 7.
Answer:
0

Question 31.
If α, β are the zeroes of the polyno¬mial x² – x – 6, then find α²β²
Answer:
36
Explanation:
αβ = \(\frac{\mathrm{c}}{\mathrm{a}}\) = – 6 ⇒ (αβ)² = (-6)² = 36

Question 32.
If ‘4’ is one of the zeroes of p(x) = x² + kx – 8, then the value of k.
Answer:
-2

Question 33.
If the polynomial p(x) = x³ – x² + 3x + k is divided by (x – 1), the remainder obtained is 3, then find the value of k.
Answer:
0
Explanation:
p(1) = 1-1 + 3 + k = 3 ⇒ k = 0

Question 34.
If one zero of the polynomial f(x) = 5x² + 13x + k is reciprocal of the other, then find the value of k.
Answer:
5
Explanation:
α = x, β = 1/x
⇒ α . β = x . 1/x = k/5 ⇒ k = 5

Question 35.
Find the number of zeroes of the po!y: nomial, whose graph is given below.
Answer:

Question 36.
Number of zeroes that can be identi¬fied by the given figure.
Answer:

Question 37.
Observe the given rectangular figure, then write its area in polynomial func-tion.

Answer:
x² – 7x – 30 = 0
Explanation:
(10 – x)(x + 3) = 10x + 30 – x² – 3x = 0
⇒ x² – 7x – 30 = 0

Question 38.
f(x) = x + 3, then find zero of f(x).
Answer:
-3

Question 39.
One zero of the polynomial 2x² + 3x + k is 1/2, then find k.
Answer:
-2
Explanation:
im 19
α + β + γ = \(\frac{-b}{a}\) = -3
\(\sqrt{5}-\sqrt{5}+\gamma\) = -3
γ = -3

Question 40.
Find factors of x² + x(a + b) + ab.
Answer:
x + a and x + b

Question 41.
f(x) = 4x² + 4x – 3, then find f(\(\frac{-3}{2}\))
Answer:
0

Question 42.
If \(\sqrt{5}\) and \(-\sqrt{5}\) are two zeroes of the polynomial x³ + 3x² – 5x – 15, then find its third zero.
Answer:
-3
Explanation:

Question 43.
If \(\sqrt{3}\) and – \(\sqrt{3}\) are the zeroes of a polynomial p(x), then find p(x).
Answer:
x² – 3

Question 44.
If ‘m’ and ‘n’ are zeroes of the polyno-mial 3x² + 11x – 4, then find the value \(\frac{\mathbf{m}}{\mathbf{n}}+\frac{\mathbf{n}}{\mathbf{m}}\)
Answer:
\(\frac{-145}{12}\)
Explanation:
3x² + 11x – 4 = 3x² + 12x – x – 4 = 0
⇒(x + 4)(3x – 1)= 0
⇒x = -4, \(\frac { 1 }{ 2 }\)

Question 45.
In the product (x + 4) (x + 2). Write the constant term.
Answer:
8

Question 46.
Find the polynomial whose zeroes are – 5 and 4.
Answer:
x² + x – 20

Question 47.
Flnd product of zeroes of 3x² = 1
Answer:
\(-\frac{1}{3}\)

Question 48.
Find the sum of the zeroes of x² + 7x + 10.
Answer:
-7

Question 49.
If one root of the polynomial
fix) = 5x² + 13x + k is reciprocal of the other, then find ‘k’.
Answer:
5

Question 50.
f(x) = 3x – 2. then find zero of f(x).
Answer:
\(\frac{2}{3}\)

Question 51.
Write the zeroes from the below graph.

Answer:
-2, 0 and 2

Question 52.
Make a quadratic polynomial whose zeroes are 5 and -2.
Answer:
x² – 3x – 10.

Question 53.
If α, β are (1w zero ai polynomial
f(x) = x² – p(x + 1) – c then find (α + 1)(β + 1).
Answer:
1 – c
Explanation:
α + β = \(\frac{-b}{a}=\frac{+p}{1}\) = +p
αβ = \(\frac{c}{a}=\frac{-p-c}{1}\) = -(p+c)
(α + 1) (β + 1) = αβ + α + β + 1
= -p – c + p + 1
= 1 – c

Question 54.
Write number of constant polynomial x² + 7x – 7.
Answer:
1

Question 55.
Write the quadratic polynomial, whose sum and product of zeroes are 1 and – 2 respectively.
Answer:
x² – x – 2

Question 56.
Find the product of the zeroes of x³ + 4x² + x – 6.
Answer:
6

Question 57.
p(x) = \(\frac{x+1}{1-x}\) , then find p(0).
Answer:
1

Question 58.
Write the maximum number of zeroes that a polynomial of degree 3 can have.
Answer:
Three

Question 59.
If x + 2 is a factor of x² + ax + 2 b and a 4 b = 4, then find Answer:
Answer:
3
Explanation:
(- 2)² – 2a + 2b = 0
⇒ 4 – 2a + 2b = 0
⇒ -a + b = -2

⇒ b = 1,a = 4- b = 4 – 1 = 3.

Question 60.
The graph of ax + b represents which type of polynomial ?
Answer:
linear polynomial.

Question 61.
If ‘1’ is the zero of the quadratic poly- nomlal x² + kx – 5, then find the value ofk.
Answer:
4
Explanation:
1 + k – 5 = 0 ⇒ k = 4.

Question 62.
Find a quadratic polynomial, the sum of whose zeroes is W and product of zero is 3.
Answer:
x² – 9

Question 63.
If y = p(x) is represented by the given pqih, then find die number of zeroes.

Answer:
4

Question 64.
If α + β = 0, αβ = \(\sqrt{3}\), then write the quadratic polynomial.
Answer:
x² + \(\sqrt{3}\)

Question 65.
Find the degree of the polynomial ax⁴ + bx³ + cx² + dx + e.
Answer:
4.

Question 66.
If α, β, γ are the zeroes of the polyno- mid f(x) = x³ – px² + qx – r, then find
\(\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}\)
Answer:
\(\frac{\mathbf{p}}{\mathrm{r}}\)
Explanation:
\(\frac{1}{\alpha \beta}+\frac{1}{\beta \gamma}+\frac{1}{\gamma \alpha}=\frac{\alpha+\beta+\gamma}{\alpha \beta \gamma}=\frac{\frac{p}{1}}{\frac{r}{1}}=\frac{p}{r}\)

Question 67.
If α and β are the zeroes of the poly¬nomial p(x) = 2x² + 5x + k satisfying the relation α² + β² + αβ = \(\frac{21}{4}\), then find k.
Answer:
2
Explanation:
α² + β² + αβ

Question 68.
If p(t) = t³ , then find p(- 2).
Answer:
– 9

Question 69.
If the polynomial f(x) = ax³ – bx – a is divisible % the polynomial g(x) = x² + bx + c, then find ah.
Answer:
1

Question 70.
Write degree of a linear polynomial.
Answer:
1

Question 71.
If the sum off the zeroes off die polyno¬mial fix) = 2x³ – 3kx² + 4x – 5 is 6, then find k.
Answer:
4.

Question 72.
If p and q are the zeroes of the poly¬nomial t² – 4t + 3, then find \(\frac{1}{p}+\frac{1}{9}-2 p q+\frac{14}{3}\)
Answer:
0
Explanation:
t² – 3t – t + 3
= t(t – 3) – 1(t-3)
= (t – 3) (t – 1)
t = 3,1
p = 3, q = 1

Question 73.
If the inudnct of zeroes rf9x2+3x + p is 7, then find p”.
Answer:
63

Question 74.
If α, β, γ are the zeroes of the polyno-mial f(x) = ax³ + hx² + cx + d, then find α² + β² + γ²
Answer:
\(\frac{b^{2}-2 a c}{a^{2}}\)

Question 75.
p(x) = x² + 5x + 6, then find zeroes of p(x).
Answer:
-2,-3.

Question 76.
If one of the zeroes of the quadratic polynomial ax² + bx + c is ‘0’, then find the other zero.
Answer:
-b

Question 77.
α = a – b, β = a 4 b, then make the quadratic polynomial.
Answer:
x² – 2ax + a² – b²

Question 78.
Find the quadratic polynomial whose zeroes are \(4+\sqrt{5}\) and \(4-\sqrt{5}\)
Answer:
x² – 8x + 11

Question 79.
Find die remainder when
3x³ + x² + 2x + 5 is divided by x² + 2x + 1.
Answer:
9x + 10

Question 80.
What must be subtracted or added to p(x) = 8x⁴ + 14x³ – 2x² + 8x – 12, so that 4x² + 3x – 2 is a factor off p(x) ?
Answer:
15x – 14

Question 81.
Prepare a quadratic polynomial whose zeroes are \(\sqrt{15}\) and \(-\sqrt{15}\).
Answer:
x² – 15

Question 82.
Divide(x³ – 6x² + 11x – 12) by (x² – x + 2)r then find quotient.
Answer:
x – 5

Question 83.
If -1 is a zero of the polynomial f(x) = x² – 7x – 8, then find the other zero.
Answer:
8

Question 84.
If a > 0, dim draw the shape off ax² + bx + c = 0.
Answer:

Question 85.
If 2x + 3 is a factor of 2x³ – x – b + 9x², then find the value erf b.
Answer:
-15

Question 86.
If the order erf ax⁵ + 3x⁴ + 4x³ + 3x² + 2x + 1 is 4, then find Answer:
Answer:
0
Explanation:
If a = 0, then given equation order becomes 4.

Question 87.
Find the zeroes erf the polynomial p(x) = 4x² – 12x + 9.
Answer:
\(\frac{3}{2}, \frac{3}{2}\)

Question 88.
If α, β, γ are roots erf a cubic polynomial ax³ + bx² + cx + d, then find αβ + βγ + γα
Answer:
\(\frac{\mathbf{c}}{\mathbf{a}}\)

Question 89.
If one of the zeroes of the quadratic polynomial f(x) = 14x² – 42k²x – 9 is negative of the other, then find k.
Answer:
0
Explanation:
Let α = x, β = – x
α + β = \(-\frac{b}{a}\)
x – x = \(\frac{-\left(-42 \mathrm{k}^{2}\right)}{14}\)
⇒ + 3k² = 0
⇒ k² = \(\frac{0}{3}\) = 0 ⇒ k = 0

Question 90.
p(x) = 4x² + 3x – 1, then find P(\(\frac{1}{4}\))
Answer:
0

Question 91.
Find the sum of zeroes of the polynomial 3x² – 8x + 1 = 0.
Answer:
\(\frac{8}{3}\)
Explanation:
3x² – 8x + 1 = 0
Sum of zeroes = \(-\frac{b}{a}\)
= \(\frac{-(-8)}{3}=\frac{8}{3}\)

Question 92.
If the product of zeroes of the polynomial f(x) = ax³ – 6x² + 11x – 6 is 4, then find Answer:
Answer:
\(\frac{3}{2}\)
Explanation:
αβγ = \(\frac{-\mathrm{d}}{\mathrm{a}}\) = 4
\(\frac{-(-6)}{a}\) = 4 ⇒ 4a = 6 ⇒ a = \(\frac{6}{4}=\frac{3}{2}\)

Question 93.
Name a polynomial of degree ‘3’.
Answer:
Cubic polynomial.

Question 94.
Is the below graph represents a polynomial ?

Answer:
No, it is not a polynomial.

Question 95.
Find the quotient when x⁴ + x³ + x² – 2x – 3 is divided by x² – 2.
Answer:
x² + x + 3

Question 96.
If α, β, γ are roots of a cubic polyno-mial ax³ + bx² + cx + d, then find α + β + γ
Answer:
\(\frac{-b}{a}\)

Question 97.
If one zero of the quadratic polynomial 2x² + kx – 15 is 3, then find the other zero.
Answer:
\(\frac{-5}{2}\)

Question 98.
If α, β, γ are the zeroes of the polynomial f(x) = ax³ + bx² + cx + d, then
find \(\frac{1}{\alpha}+\frac{1}{\beta}+\frac{1}{\gamma}\)
Answer:
\(-\frac{c}{d}\)

Question 99.
The graph

represents, which type of polynomial ?
Answer:
Cubic polynomial.

Question 100.
Write the product and sum of the zeroes of the quadratic polynomial ax² + bx + c respectively.
Answer:
\(\frac{c}{a}, \frac{-b}{a}\)

Question 101.
Write the number of zeroes of the poly¬nomial in the given graph.

Answer:
0

Question 102.
If f(x) = ax² + bx + c has no real zeroes and a + b + c < 0, then write the condition.
Answer:
c < 0

Question 103.
Which type of polynomial ax² + bx + c?
Answer:
Quadratic polynomial.

Question 104.
If one zero of the polynomial
f(x) = (k² + 4)x2 + 13x + 4k is recip¬rocal of the other, then find k.
Answer:
2
Explanation:
\(x \cdot \frac{1}{x}=\frac{4 k}{k^{2}+4}\)
⇒ k² + 4 = 4k
=» k² – 4k + 4 = 0
⇒ (k – 2)² = 0
⇒ k = 2

Question 105.
Write the number of zeroes of the poly-nomial function p(x), whose graph is given below.

Answer:
3

Question 106.
If two zeroes of x³ + x² – 5x – 5 are \(\sqrt{5}\) and \(-\sqrt{5}\) then find its third zero.
Answer:
-1

Question 107.
If one zero of the quadratic polynomial x² – 5x – 6 is 6, then find the other zero.
Answer:
– 1

Question 108.
Find the degree of the polynomial \(a_{0} x^{n}+a_{1} x^{n-1}+a_{2} x^{n-2}+\ldots .+a_{n} x^{n}\)
Answer:
n

Question 109.
If α and β are the zeros of the polyno-mial f(x) = x² + px + q, then find a
polynomial having \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) as its a p
zeroes.
Answer:
qx² + px + 1

Question 110.
If both the zeroes of a quadratic poly-nomial ax² + bx + c are equal and opposite in sign, then find ‘b’.
Answer:
0

Question 111.
Write the number of zeroes of the poly-nomial in the given graph.

Answer:
1

Question 112.
Find the sum and product of the ze¬roes of polynomial x² – 3 respectively.
Answer:
0,-3

Question 113.
If a < 0, then draw the shape of ax² + bx + c = 0.
Answer:

Question 114.
From the graph write the number of zeroes of the polynomial.

Answer:
2

Question 115.
If α and β are zeroes of the polynomial p(x) = x² – 5x + 6, then find the value of α + β – 3αβ.
Answer:
– 13
Explanation:
α + β = \(\frac{-b}{a}=\frac{-(-5)}{1}\) = 5
αβ = \(\frac{\mathrm{c}}{\mathrm{a}}\) = 6
= α + β – 3αβ = 5 – 3(6)
= 5- 18 = – 13.

Question 116.
What should be subtracted from the polynomial x² – 16x + 30 so that 15 is the zero of the resulting polynomial ?
Answer:
15

Question 117.
Find the number of zeroes lying be-tween – 2 and 2 of the polynomial f(x) whose graph is given below.

Answer:
2

Question 118.
If the zeroes of a quadratic polynomial are equal in magnitude but opposite in sign, then write the relation be¬tween zeroes.
Answer:
Sum of its zeroes is 0.

Question 119.
Find where the graph of the polyno¬mial f(x) = 2x – 5 is a straight line which intersects the x – axis.
Answer:
[\(\frac{5}{2}\),0]

Question 120.
If α,β are the zeroes of the polyno¬mial f(x) = ax² + bx + c, then find \(\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}=\)
Answer:
\(\frac{b^{2}-2 a c}{c^{2}}\)

Choose the correct answer satis¬fying the following statements.

Question 121.
Statement (A): (2 – \(\sqrt{3}\))is °ne zero of the quadratic polynomial, then other zero will be (2 + \(\sqrt{3}\)).
Statement (B) : Irrational zeroes (roots) always occurs in pairs.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
As irrational roots / zeroes always oc¬curs in pairs therefore, when one, zero
is 2 – √3 , then other will be 2 + √3
So, both A and B are correct and B explains A.
Hence, (i) is the correct option.

Question 122.
Statement (A) : If both zeroes of the quadratic polynomial x² – 2kx + 2 are equal in magnitude but opposite in sign, then value of k is \(\frac { 1 }{ 2 }\).
Statement (B) : Sum of zeroes of a quadratic polynomial ax² + bx + c is \(\frac{-b}{a}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation:
As the polynomial is x² – 2kx + 2 and
its zeros are equal but opposition sign.
∴ Sum of zeroes = 0 = \(\frac{-(-2 \mathrm{k})}{1}\) = 0
⇒ 2k = 0 ⇒ k = 0
So, A is incorrect but B is correct.
Hence, (iii) is the correct option.

Question 123.
Statement (A): p(x) = 14x³ – 2x² + 8x⁴ + 7x – 8 is a polynomial of degree 3. Statement (B) : The highest power of x in any polynomial p(x) is the degree of the polynomial.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation:
The highest power of x in the polyno¬mial p(x) = 14x³ – 2x² + 8x⁴ + 7x – 8 is 4.
∴ Degree of p(x) is 4.
So, A is incorrect but B is correct.
Hence, (iii) is the correct option.

Question 124.

Statement (A) : The graph y = f(x) is shown in figure, for the polynomial f(x). The number of zeroes of f(x) is 4.
Statement (B): The number of zero of polynomial f(x) is the number of point of which f(x) cuts or touches the axes.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)
Explanation:
As the number of zero of polynomial f(x) is the number of {mints at which f(x) cuts (intersects) the x-axis and number of zero in the given figure is 4. So A is correct but B is incorrect Hence, (ii) is the correct option.

Question 125.
Statement (A) : The sum and product of the zeroes of a quadratic polynomial
are – \(\frac { 1 }{ 4 }\) and \(\frac { 1 }{ 4 }\) respectively. Then the
quadratic polynomial is 4x² + x + 1.
Statement (B): The quadratic polyno-mial whose sum and product of zeroes are given is x² – (sum of zeroes)x + product of zeroes.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false,
Answer:
(i)
Explanation:
Sum of zeroes = –\(\frac { 1 }{ 4 }\) and
Product of zeroes = \(\frac { 1 }{ 4 }\)
∴ Quadratic polynomial be

∴ Quadratic polynomial be 4x² + x + 4.
So, both A and B are correct and B explains A. Hence, (i) is the correct option.

Question 126.
Statement (A) : If α, β are zeroes of x² – 3x + p and 2α + 3β = 15, then p – 54.
Statement (B) : If α, β are the zeroes of the polynomial ax² + bx + c, then
α + β = \(\frac{-b}{a}\) and αβ = \(\frac{c}{a}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)

Question 127.
Statement (A) : A quadratic polyno¬mial having 4 and – 2 as zeroes is 2x² – 4x – 16.
Statement (B): The quadratic polyno-mial having a and (3 as zeroes is given by p(x) = x² – (α + β)x + αβ
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 128.
Statement (A) : The polynomial p(x) = x³ + x has one real zero.
Statement (B) : A polynomial of nth degree has at most n zeroes.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)

Question 129.
Statement (A): If α, β, γ are the zeroes of x³ – 2x² – qx – r and α + β = 0, then 2q= r.
Statement (B) : If α, β, γ are the ze¬roes of ax³ + bx² + cx + d, then
α + β + γ = \(\frac{-b}{a}\), αβ, βγ, γα = \(\frac{c}{a}\), αβγ = \(\frac{-d}{a}\)
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation:
Clearly, (B) is true, [standard result] cc + p + y = -(-2) = 2 ⇒ 0 + y = 2 ‘ Y – 2 ‘
α + β + γ = – (- 2) = 2=
o + γ = 2
γ = 2
αβγ = -(-r) = r
∴ αβ(2) = r
⇒ αβ = \(\frac{r}{2}\)
⇒ αβ + βγ + γα = -q
⇒\(\frac{r}{2}\) + γ(α+β) = -q
⇒\(\frac{r}{2}\) + 2(0) = -q
⇒ r = 2q
∴ (A) is false
Hence, (iii) is the correct option.

Question 130.
Statement (A) : If one zero of polyno¬mial p(x) = (k² + 4)x² + 13x + 4k is reciprocal of other, then k = 2.
Statement (B) : If x – a is a factor of p(x), then p(α) = 0 i.e., a is a zero of p(x).
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)
Explanation:
(B) is true
Let α, 1/α be the zeroes of p(x), then
\(\alpha \cdot \frac{1}{\alpha}=\frac{4 \mathbf{k}}{\mathbf{k}^{2}+4} \Rightarrow \mathbf{1}=\frac{4 \mathbf{k}}{\mathbf{k}^{2}+4}\)
∴ k² -4k + 4 = 0 ⇒ (k- 2)² = 0
∴ k = 2
∴ (A) is true. So, (i) is correct option.

Question 131.
Statement (A) : The polynomial x⁴ + 4x² + 5 has four zeroes. Statement (B) : If p(x) is divided by (x – k), then the remainder = p(k).
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(iii)
Explanation:
(B) is true by remainder theorem. Again, x⁴ + 4x² + 5
= (x² + 2)² + 1 > 0 for all x.
∴ Given polynomial has no zeroes.
∴ (A) is not true.
Hence, (iii) is the correct option.

Question 132.
Statement (A) : x³ + x has only one real zero.
Statement (B) : A polynomial of nth degree must have ‘n’ real zeroes.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)
Explanation:
(B) is false [v a polynomial of n<sup<th degree has at most x zeroes]
Again, x³ + x = x (x² + 1) which has only one real zero (x = 0)
[∵ x² + 1 ≠ 0 for all x ∈ R]
(A) is true.
Hence, (ii) is the correct option.

Question 133.
Statement (A) : If 2, 3 are the zeroes of a quadratic polynomial, then poly¬nomial is x² – 5x + 6.
Statement (B) : If a, P are the zeroes of a monic quadratic polynomial, then polynomial is x² – (a + p)x + ap.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 134.
Statement (A): Degree of a zero poly-nomial is not defined.
Statement (B) : Degree of a non-zero constant polynomial is ‘0’.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(ii)

Question 135.
Statement (A) : Zeroes of f(x) = x³ – 4x – 5 are 5, – 1.
Statement (B): The polynomial whose zeroes are \(2+\sqrt{3}, 2-\sqrt{3}\) is x² – 4x + 7.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Question 136.
Statement (A) : x² + 4x + 5 has two zeroes.
Statement (B) : A quadratic polyno¬mial can have at the most two zeroes.
i) Both A and B are true.
ii) A is true, B is false.
iii) A is false, B is true.
iv) Both A and B are false.
Answer:
(i)

Read the below passages and an¬swer to the following questions.

If α, β are the zeroes of the quadratic polynomial f(x) = ax² + bx + c, then
α + β = \(\frac{-b}{a}\), αβ = \(\frac{c}{a}\)

If α, β are the zeroes of the quadratic polynomial f(x) = x² – px + q, then find \(\frac{1}{\alpha}+\frac{1}{\beta}\)
Answer:
\(\frac{p}{q}\)
Explanation:
α + β = p, αβ = q
∴ \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}=\frac{p}{q}\)

Question 138.
If α, β are the zeroes of the quadratic polynomial fix) = x² + x – 2, then find \(\left(\frac{1}{\alpha}-\frac{1}{\beta}\right)^{2}\)
Answer:
\(\frac{9}{4}\)
Explanation:
α + β = -1, αβ = -2

Question 139.
If α, β are the zeroes of the quadratic polynomial fix) = x² – 5x + 4, then
find \(\frac{1}{\alpha}+\frac{1}{\beta}-2 \alpha \beta\)
Answer:
–\(\frac{27}{4}\)

If α, β, γ are the zeroes of ax³ + bx² + cx + d, then \(\Sigma \alpha=\frac{-b}{a}\), \(\Sigma \alpha \beta=\frac{\mathbf{c}}{\mathbf{a}}, \Sigma \alpha \beta \gamma=\frac{-\mathbf{d}}{\mathbf{a}}\)
Explanation:
α + β = 5, αβ = 4

Question 140.
If α, β, γ are the zeroes of x³ – 5x² – 2x + 24 and ap = 12, then
find αβ = 12, then find ‘γ’
Answer:
-2

Question 141.
If a – b, a, a + b are the roots of x³ – 3x² + x + 1, then find a + b².
Answer:
3

Question 142.
If two zeroes of the polynomial x³ – 5x² – 16x + 80 are equal in magni¬tude but opposite in sign, then find ze¬roes.
Answer:
4,-4, 5.

Manow says that the order of the polynomial (x² – 5) (x³ + 1) is 6.

Question 143.
Do you agree with Manow ?
Answer:
No.

Question 144.
Which mathematical concept is used to judge Manow ?
Answer:
Polynomial.

Question 145.
How much the actual order of given problem ?
Answer:
Degree is 5.

The length of a rectangle is ’5’ more than its breadth.

Question 146.
Express the information in the form of polynomial.
Answer:
(x + 5 + x) = 2x + 5.

Question 147.
Find the perimeter of the rectangle given above.
Answer:
(4x + 10)m

Question 148.
To solve this given problem which mathematical concept was used by you?
Answer:
Polynomial.

Question 149.
Write the correct matching option.

Answer:
A – (ii), B – (iv).

Question 150.
Write the correct matching option.

A – (iii), B – (i).

Question 151.
Write the correct matching option.

Answer:
A – (iv), B – (iii).

Question 152.
Write the correct matching option.

Answer:
A – (ii), B – (i).

Question 153.
Write the correct matching option.

Answer:
A – (iii), B – (i).

Question 154.
Write the correct matching option.

Answer:
A – (iii), B – (v).

Question 155.
Write the correct matching option.

Answer:
A – (iv), B – (i).

Question 156.
If α, β, γ are the zeroes of the polyno-mial px³ + qx² + rx + s then, which of the following matching is correct ?

a) A(i), B(ii), C(iii)
b) A(ii), B(iii), C(i)
c) A(iii), B(i), C(ii)
d) A(ii), B(i), C(iii)
Answer:
(b)

Question 157.
What is the zero of the polynomial 3x – 2 ?
Solution:
f(x) = 3x – 2; f(x) = 0
3x – 2 = 0 ⇒ 3x = 2
⇒ x = 2/3

Question 158.
Write the polynomial in variable ‘x’ whose zero is \(\frac{-k}{a}\).
Solution:
x – \(\frac{-k}{a}\) = 0 ⇒ x + \(\frac{k}{a}\) = 0
⇒ ax + k = 0
∴ ax + k = 0 is a polynomial with degree T in variable ‘x’ whose zero is \(\frac{-k}{a}\).