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Andhra Pradesh SSC Class 10 Solutions For Maths – Polynomials (English Medium)
These Solutions are part of AP SSC Class 10 Solutions for Maths. In this article, we have provided Andhra Pradesh SSC Class 10 Solutions For Maths Chapter 3 Polynomials.
Exercise 3.1:
Question 1:
(a) If p(x) = 5x7 – 6x5 + 7x-6, find
(i) coefficient of x5
(ii) degree of p(x)
(iii) constant term.
Solution :
a.
i. Coefficient of x5 = -6
ii. Degree of p(x) = 7
iii. Constant term = -6
AP SSC 10th Class Textbook Solutions
b.
(i). p(x)= x2-2x-3
1. coefficient of x2 =12.
2. Degree of p(x) = 2
3. Constant term = -3
(ii). p(x)=3x8+4x5-1
1. Value of p(1)=3(1)8+4(1)5-1
i. =3+4-1
ii. =6
2. Constant term = -1
3.Coefficient of x5 = 4
iii. p(x)= 2x3+3x+5
1) Degree of p(x)=3
2) Value of p(2)= 2(2)3+3(2)+5
=16+6+5 = 27
3) Degree of Constant term=0
Question 2:
State which of the following statements are true and which are false? Give reasons for your choice.
- The degree of the polynomial \(\sqrt { 2 } { x }^{ 2 }-3x+1\quad is\quad \sqrt { 2 }.\)
- The coefficient of x2 in the polynomial p(x) = 3x3 – 4x2 + 5x + 7 is 2.
- The degree of a constant term is zero.
- \(\frac { 1 }{ { x }^{ 2 }-5x+6 }\) is a quadratic polynomial.
- The degree of a polynomial is one more than the number of terms in it.
Solution :
- False ( √2 is a coefficient of x2 not a degree)
- False (coefficient of x2 is -4)
- True (For any constant term, degree is zero)
- False (It is not a polynomial at all)
- False (Degree of polynomial is not related to number of terms)
Question 3:
If p( t) = t3 – 1, find the values of p(1), p(-1), p(0), p(2), p(-2).
Solution :
Given, p(t)= t3-1
p(1) = (1)3-1=0
p(-1)= (-1)3-1= -1-1= -2
p(0) = (0)3-1 = -1
p(2)= (2)3-1=8-1= 7
p(-2) = (-2)3-1 = -8-1=-9
Question 4:
Check whether -2 and 2 are the zeroes of the polynomial x4 – 16
Solution :
Given, p(x)= x4 – 16
p(-2) = (-2)4 – 16 = 16 – 16 = 0
p(2) = (2)4 – 16 = 16 – 16 = 0
Hence, -2 and 2 are the zeroes of the polynomial x4 – 16.
Question 5:
Check whether 3 and -2 are the zeroes of the polynomial p(x) when p(x) = x2 – x – 6.
Solution :
Given, p(x)= x2– x – 6
p(3) = (3)2 – (3) – 6 = 9 – 3 – 6 = 0
p(-2) = (-2)2 – (-2) -6 = 4 + 2 – 6 = 0
Hence, 3 and -2 are the zeroes of the polynomial p(x) = x2 – x – 6.
Exercise 3.2:
Question 1:
The graphs of’v = p(x) are given in the figure below, for some polynomials p(x). In each case, find the number of zeroes of p(x).
Solution :
case, find the number of zeroes of p(x).
- The number of zeroes is 0 as the graph does not intersect x-axis at any point.
- The number of zeroes is 1 as the graph intersects x-axis at one point only.
- The number of zeroes is 3 as the graph intersects x-axis at three points.
- The number of zeroes is 2 as the graph intersects x-axis at two points.
- The number of zeroes is 4 as the graph intersects x-axis at four points.
- The number of zeroes is 3 as the graph intersects x-axis at three points.
Question 2:
Find the zeroes of the given polynomials.
- p(x) = 3x
- p(x) = x2 + 5x + 6
- p(x) = (x+2) (x+3)
- p(x) = x4 – 16
Solution :
- p(x)= 3x is a linear polynomial. It has only one zero.
To find zeroes,
Let p(x)=0
So,3x=0
∴ x=0 - p(x)= x2+5x+6 is a quadratic polynomial. It has at most two zeroes.
To find zeroes,
Let p(x)=0
x2+5x+6=0
(x+2)(x+3)=0
x=-2 or x=-3
∴The zeroes of the polynomial are -2 and -3. - p(x)= (x+2)(x+3) is a quadratic polynomial. It has at most two zeroes.
To find zeroes, let p(x)=0
(x+2)(x+3) =0
x=-2 or x=-3
∴The zeroes of the polynomial are -2 and -3. - p(x) = x4-16
To find zeroes,
Let p(x)= 0
x4-16=0
(x2-4)(x2 +4)=0
Discarding x2=-4,
x2=4
x=±√4
∴x=-2 or x=2
∴The zeroes of the polynomial are -2 and 2.
Question 3:
Draw the graphs of the given polynomial and find the zeroes. lustily the answers,
Question 3(i):
p(x) = x2 – x – 12
Solution :
Question 3(ii):
p(x) = x2 – 6x + 9
Solution :
Question 3(iii):
p(x) = x2 – 4x + 5
Solution :
Question 3(iv):
p(x) = x2 + 3x – 4
Solution :
Question 3(v):
p(x) = x2 -1
Solution :
Question 4:
Why are 1/4 and -1 zeroes of the polynomials p(x) = 4x2 + 3x- 1?
Solution :
Exercise 3.3:
Question 1:
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
Question 1(i):
x2 – 2x – 8
Solution :
We can write:
x2-2x-8= (x+2)(x-4)
so, the value of x2-2x-8 is zero when (x+2)=0 or (x-4)=0,i.e. when x=-2 or x=4
Therefore, the zeroes of x2-2x-8 are -2 and 4.
Now,
sum of the zeroes= (-2)+4=2
Question 1(ii):
4s2 – 4s + 1
Solution :
Question 1(iii):
6x2 – 3 – 7x
Solution :
Question 1(iv):
4u2 + 8u
Solution :
We can write:
4u2+8u= (2u+4)(2u)
so, the value of 4u2+8 is zero when (2u+4)=0 or (2u)=0,
i.e. when u=-2 or u=0
Therefore, the zeroes of 4u2+8u are -2 and 0.
Now,
Sum of the zeroes=(-2)+0=-2
Product of the zeroes= -2×0= 0
To verify the relation between zeroes and coefficients we have to prove
Question 1(v):
t2 – 15
Solution :
Question 1(vi):
3x2 – x – 4
Solution :
Question 2:
Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes respectively.
Question 2(i):
\(\frac { 1 }{ 4 } ,-1\)
Solution :
Let the quadratic polynomial be ax2+bx+c, α ≠ 0and its zeroes be α and β.
We have
If we take a=4 ,then b=-1 and c=-4
So, one quadratic polynomial which fits the given condition is 4x2-x-4.
Question 2(ii):
\(\sqrt { 2 } ,\frac { 1 }{ 3 }\)
Solution :
Question 2(iii):
\(0,\sqrt { 5 }\)
Solution :
Question 2(iv):
1,1
Solution :
Question 2(v):
\(-\frac { 1 }{ 4 } ,\frac { 1 }{ 4 } \)
Solution :
Question 2(vi):
4,1
Solution :
Question 3:
Solution :
Question 4:
Verify that 1,-1 and -3 are the zeroes of the cubic polynomial x3 + 3x2 – x – 3 and check the relationship between zeroes and the coefficients.
Solution :
Comparing the given polynomial with ax3+bx2+cx+d, we get a=1, b=3,c =-1 and d=-3.
P(1)= (1)3+3(1)2-1-3=1+3-1-3=0
P(-1)= (-1)3+3(-1)2-(-1)-3 = -1+3+1-3= 0
P(-3)= (-3)3+3(-3)2-(-3)-3 = -27+27+3-3 = 0
Therefore, 1, -1 and -3 are the zeroes of x3+3x2-x-3.
So, we take α=1and β=-1 and γ =-3
Now,
Exercise 3.4:
Question 1:
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
Question 1(i):
p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2
Solution :
Dividend= x3-3x2+5x-3 Divisor= x2-2
We stop here since degree of the remainder is less than the degree of (x2-2) the divisor.
So, quotient= x-3, remainder=7x-9
Now,
Dividend=Divisor x Quotient + Remainder
= (x2-2) x (x-3) +(7x-9)
= x3– 3x2– 2x + 6 +7x -9
= x3-3x2+5x-3
Thus, division algorithm is verified.
Question 1(ii):
p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x
Solution :
dividend= x4-3x2+4x+5 divisor= x2-x+1
We stop here since degree of the remainder is less than the degree of (x2-x+1) the divisor.
So, quotient= x2+x-3, remainder=8
Now,
Dividend=Divisor x Quotient + Remainder
= (x2+1-x)x(x2+x-3)+8
= x4-3x2+4x+5
Thus, division algorithm is verified.
Question 1(iii):
p(x) = x4 – 5x + 6, g(x) = 2 – x2
Solution :
dividend= x4-5x+6 divisor= -x2+2
We stop here since degree of the remainder is less than the degree of (-x2+2) the divisor.
So, quotient= -x2-2, remainder=-5x+10
Now,
Dividend=Divisor x Quotient + Remainder
= (-x2+2) x (-x2-2) +(-5x+10)
= x4-5x+6
Question 2:
Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
- t1 – 3, 2t4 + 3t3 – 2t2 – 9t – 12
- x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2
- x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1
Solution :
- dividend= 2t4-3t3-2t2-9t-12
divisor= t2-3
Yes, first polynomial is a factor of the second polynomial
- dividend=3x4+5x3-7x2+2x+2
divisor = x2+3x+1
Yes, first polynomial is a factor of the second polynomial
first polynomial is a factor of the second polynomial
Question 3:
Obtain all other zeroes of 3x4 + 6x3 – 2x- 10x- 5, if two of its zeroes are
\(\sqrt { \frac { 5 }{ 3 } } and-\sqrt { \frac { 5 }{ 3 } }.\)
Solution :
Question 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).
Solution :
Dividend = x3 – 3x2 + x + 2
Quotient = x – 2
Remainder = -2x + 4
Now,
Dividend= Divisor x Quotient+ Remainder.
x3-3x2+ x + 2 = Divisor x (x – 2) + (-2x + 4)
x3 – 3x2 + x + 2 – (-2x + 4)= divisor x (x – 2)
x3 – 3x2+ x + 2 + 2x – 4 = divisor x (x – 2)
x3 – 3x2 + 3x + 2= divisor x (x – 2)
Hence, divisor=x2-x+1
Question 5:
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
- deg p(x) = deg q(x)
- deg q(x) = deg r(x)
- deg r(x) = 0
Solution :
- p(x) = 2x2 – 2x + 14 , g(x) = 2, q(x) = x2 – x + 7, r(x) = 0
- p(x) = x3 + x2 + x + 1, g(x) = x2 – 1, q(x) = x + 1, r(x) = 2x + 2
- p(x) = x3 + 2x2 – x + 2, g(x) = x2 – 1, q(x) = x + 2, r(x) = 4.
Hope given Andhra Pradesh SSC Class 10 Solutions For Maths chapter 3 Polynomials are helpful to complete your math homework.
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