These AP 10th Class Maths Chapter Wise Important Questions Chapter 3 Polynomials will help students prepare well for the exams.
AP State Syllabus 10th Class Maths 3rd Lesson Important Questions and Answers Polynomials
Question 1.
If p(x) = x2 – 5x – 6; find the value of p(3).
Solution:
p(x) = x2 – 5x – 6 (given) .
p(3) = 32 – 5(3) – 6
= 9 – 15 – 6
= 9 – 21
= -12
So p(3) = – 12.
Question 2.
Write two more pdjmmilak and cre¬ate two questions for each of them.
Solution:
i) x2 – 5x + 6 (polynomial (i)}
Questions:
1) What is the order of the polynomial ?
2) What are the zeroes of the polynomial?
ii) x2 – (4x) (polynomial (ii))
Questions:
1) How many zeroes are there to this polynomial ?
2) Do this polynomial has ’0’ as its zero value ?
Question 3.
If p(y) = y3 – 1, then verify that 1, -1 are the zeroes of p(y) or not.
Solution:
p(y) = y3 – 1
p(1) = 13 – 1 = 1 – 1 = 0
∴ ‘1’ is the zero of p(y).
p(-1) = (-1) – 1 = -1 – 1 = -2
∴ ‘-1’ is not zero of p(y).
Question 4.
For what value of k, is 3 a zero (tf the polynomial 2x2 + x + k
Solution:
Since 3 is a zero of the polynomial p(x) = 2x2 + x + k,
p(3) = 0
∴ p(3) = 2(3)2 + 3 + k = 0
⇒ 21 + k = 0
⇒ k = -21
Question 5.
Find the remainder when
7x5 – 3x2 + 5x – 2 is divided by x + 2.
Solution:
∴ Remainder = – 80
Question 6.
Draw a rough graph of x – y = 0.
Solution:
| x | y | (x, y) |
| 0 | 0 | (0,0) |
| 1 | 1 | (1, 1) |
| -1 | -1 | (-1, -1) |
| -2 | -2 | (-2, -2) |
| 2 | 2 | (2, 2) |
Question 7.
Find the quadratic polynomial, whose zeroes are \(\frac { 2 }{ 3 }\) and 2.
Solution:
Let the quadratic polynomial be
ax2 + bx + c, a ≠ 0 and its zeroes be α and β.
Here α = \(\frac { 2 }{ 3 }\) and β = 2.
Sum of the zeores = α + β = \(\frac { 2 }{ 3 }\) + 2 = \(\frac { 8 }{ 3 }\)
Product of the zeroes = αβ = \(\frac { 2 }{ 3 }\)(2) = \(\frac { 4 }{ 3 }\)
∴ The quadratic polynomial
ax2 + bx + c is = [x2 – (α + β)x + αβ]
= [x2 – \(\frac { 8 }{ 3 }\)x + \(\frac { 4 }{ 3 }\)]
The quadratic polynomial will be 3x2 – 8x + 4.
Question 8.
Find the zeroes of the quadratic poly-nomial 5x2 -4 – 8x and verify the rela-tion between the zeroes and its co-efficients.
Solution:
Given polynomial is 5x2 – 4 – 8x
= 5x2 – 8x – 4
= 5x2 – 10x + 2x – 4
= 5x(x – 2) + 2(x – 2)
= (x – 2) (5x + 2)
To find zeroes, (x – 2) (5x + 2) = 0
⇒ x – 2 = 0 or 5x + 2 = 0
⇒ x = 2 or x = \(-\frac { 2 }{ 5 }\)
Question 9.
Write the quadratic polynomial, whose sum of zeroes is —3 and sum of the i of zeroes is 17.
Solution:
The general form of quadratic polynomial having α,β as its zeroes is
= x2 – (sum of zeroes)x + product of zeroes.
= x2 – (α + β) x + αβ,
here given (α + β) = -3 and
(α2 + β2) = 17
∴ (α + β)2 = α2 + β2 + 2αβ
(-3)2 = 17 + 2αβ => 2αβ = 9 – 17 = -8
∴ αβ = \(-\frac{8}{2}\) = -4
∴ The required polynomial is x2 – (-3) x + (-4) = x2 + 3x – 4 is required quadratic polynomial.
Question 10.
if p and q are the zeroes of the poly¬nomial 3x2 – 5x + 2, write the polynomial in ‘x’ whose zeroes are \(\frac { 1 }{ p }\) and \(\frac { 1 }{ q }\)
Solution:
Solution:
It is given that, p and q are zeroes of polynomial 3x2 – 5x + 2.
Sum of zeroes = p + q = \(\frac{-(-5)}{3}=\frac{5}{3}\) ………….(1)
Product of zeroes = pq = \(\frac{2}{3}\) ……………(2)
Now we have to find the polynomial zeroes of which are \(\frac { 1 }{ p }\) and \(\frac { 1 }{ q }\).
Hence required polynomial is 2x2 – 5x + 3.
Question 11.
Solve the quadratic; polynomial x2 – 3x – 4 by graphical method.
Solution:
Let y = x2 – 3x – 4
For finding the points to draw y = x2 – 3x – 4
For plotting the points and drawing the graph.
Note : The limitations of the solution in the problem need not be considered and award the marks for drawing the graph.
Question 12.
Draw the graph of polynomial 4x2 + 4x – 3 and find the zeroes, using the graph.
Solution:
To find the zeroes of the polynomial 4x2 + 4x – 3 by graph method. We need to find the points that lie on the curve.
So zero values of y = 4x2 + 4x – 3 are 0.5 and -1.5
We observe the above curve passes through the points (0.5, 0) and (-1.5, 0) which are on x-axis.
Hence the zeroes of above polynomials are 0.5 and – 1.5.
Question 13.
Find the zeroes of the quadratic polynomial p(x) = x2+ x – 20 using graph.
Solution:
Given polynomial = p(x) = x2 + x – 20
List of values of p(x) :
Now let’s locate the points listed above on a graph paper and join them free hand.
Result: We observe that the graph cuts the x-axis at (4, 0) and (-5, 0).
So, the zeroes of the given polynomial are 4 and -5.
Justification : Given p(x) = x2 + x – 20
⇒ x2 + 5x – 4x – 20 = 0
⇒ x(x + 5) – 4(x + 5) = 0
⇒ (x + 5) (x – 4) = 0
⇒ x = -5 dnd x = 4
Question 14.
Find the zeroes of the quadratic polynomial p(x) = x2 – x – 2, by using the graph.
Solution:
Let y = x2 – x – 2
Zeroes of the polynomial are 2,-1.
Question 15.
Draw the graph of polynomial
p(x) = x2 – 3x + 2 and find the zeros from the graph. June 2019^
Solution:
Let y = p(x) = x2 – 3x + 2
If x = 0 then p(0) =0-0 + 2 = 2 So (0, 2)
x = 1 then p(1) = 12 + 3(1) + 2
= 1 – 3 + 2
= 0 So (1, 0)
x = 2 then p(2) = 22 – 3(2) + 2 = 4 – 6 + 2 = 0 So (2, 0)
x = 3 then p(3) = 32 – 3(3) + 2 = 9 – 9 + 2 = 2 So (3, 2)
and if x = -1 then p(-1)
= (-1)2 – 3(-1) + 2 = 1 + 3 + 2 = 6. So (-1, 6)
x = -2 then p(-2) = (-2)2 – 3 (-2) + 2 = 4 + 6 + 2= 12
So (-2, 12) that means the graph of the polyno¬mial
p(x) = x2 – 3x + 2 passes through the points.
(0,2), (1,0) (2,0) (3, 2) (-1,6) and (-2,12)
So 1 and 2 are zeros of the given poly¬nomial.
Question 16.
Find the zeroes of x2 + 8x + 15.
Solution:
x2 + 8x + 15 = x2 + 5x + 3x + 15
= x(x + 5) + 3(x + 5)= (x + 3) (x + 5)
∴ -3,-5 are zeroes of the polynomial x2 + 8x + 15.
Question 17.
How do you say that the maximum number of zeroes of x2 + 8x + 15 is 2 ?
Solution:
The order of polynomial x2 + 8x + 15 is 2. Hence the maximum number of zeroes of it is 2.
Since the maximum possible zeroes of a polynomial is equal to its order.
Question 18.
Find the zeroes of x2 + 8x + 15 using graph.
Solution:
The zeroes of a polynomial are the X-coordinates of the points. Where its graph touches X-axis. Now let us find the points to draw the graph.
So the graph of polynomial x2 + 8x + 15 passes through the points (0, 15),
(1, 24), (2, 35), (-1, 8), (-2, 3), (-3, 0) and (-4, -1), (-5, 0)
The adjacent graph intersect the X-axis at – 3 and – 5.
Hence they are zeroes of it.
Question 19.
Check whether ‘6’ becomes a zero of
x2 + 8x + 15 or not ? Give reasons.
Solution:
If p(k) = 0 then ‘k’ is a zero value of p(x)
Now p(x) = x2 + 8x + 15 then
p(6) = 62 + 8 (6) + 15
= 36 + 48 + 15 ≠ 0
As p(6) ≠ 0, ‘6’ cannot be the zero value of p(x).
Question 20.
8 times of a number is added to its square give a result – 15. Then find the number by using its quadratic equation.
Solution:
Let us consider the given number = x
Then its square = x2
8 times of it = 8x
Now adding the above two x2 + 8x
result is = – 15
x2 + 8x = -15
x2 + 8x + 15 = 0
x2 + 5x + 3x + 15 = 0
⇒ x(x + 5) + 3(x + 5) = 0
⇒ (x + 3) (x + 5) = 0
∴ x + 3 = 0 or x + 5 = 0
Then x = – 3 or – 5
Question 21.
In how many points will the graph of x2 + 8x + 15 intersect X-axis ? Why ?
Solution:
We can write x2 + 8x + 15 as x2 + 5x + 3x + 15
⇒ (x + 5) (x + 3)
So its zero values are -5, -3
So the graph of x2 + 8x + 15 intersect x-axis at two points only.
Question 22.
Express the polynomial x2 + 8x + 15 in view of variable ‘y’.
Solution:
y = x2 + 8x + 15
= x2 + 5x + 3x + 15
= x (x + 5) + 3 (x + 5)
y = (x + 3) (x + 5)
Question 23.
State the relation between coefficient of polynomials and zeroes of it.
Solution:
The polynomial is x2 -1- 8x + 15
Question 24.
Find p(3) if p(x) = x2 – 5x + 6 is given.
Solution:
p(x) = x2 – 5x + 6 (given)
then p(3) = 32 – 5(3) + 6
= 9 – 15 + 6 = 15 – 15 = 0
∴ P(3) = 0