ABCD
Basic Concepts
A polynomial is an expression consists of constants, variables and exponents. It’s mathematical form is-
anxn + an-1xn-1 + an-2xn-2 + a2x2 + a1x + a0 = 0
where the (ai)’s are constant
Degree of Polynomials
Let P(y) is a polynomial in y, then the highest #ffffcc power of y in the P(y) will be the degree of polynomial P(y).
Types of Polynomial according to their Degrees
| Type of polynomial | Degree | Form |
| Constant | 0 | P(x) = a |
| Linear | 1 | P(x) = ax + b |
| Quadratic | 2 | P(x) = ax2 + ax + b |
| Cubic | 3 | P(x) = ax3 + ax2 + ax + b |
| Bi-quadratic | 4 | P(x) = ax4 + ax3 + ax2 + ax + b |
Value of Polynomial
Let p(y) is a polynomial in y and α could be any real number, then the value calculated after putting the value y = α in p(y) is the final value of p(y) at y = α. This shows that p(y) at y = α is represented by p (α).
Zero of a Polynomial
If the value of p(y) at y = k is 0, that is p (k) = 0 then y = k will be the zero of that polynomial p(y).
Geometrical meaning of the Zeroes of a Polynomial
Zeroes of the polynomials are the x coordinates of the point where the graph of that polynomial intersects the x-axis.
Graph of a Linear Polynomial
Graph of a linear polynomial is a straight line which intersects the x-axis at one point only, so a linear polynomial has 1 degree.
Graph of Quadratic Polynomial
Case 1: When the graph cuts the x-axis at the two points than these two points are the two zeroes of that quadratic polynomial.
Case 2: When the graph cuts the x-axis at only one point then that particular point is the zero of that quadratic polynomial and the equation is in the form of a perfect square
Case 3: When the graph does not intersect the x-axis at any point i.e. the graph is either completely above the x-axis or below the x-axis then that quadratic polynomial has no zero as it is not intersecting the x-axis at any point.
Hence the quadratic polynomial can have either two zeroes, one zero or no zero. Or you can say that it can have maximum two zero only.
Relationship between Zeroes and Coefficients of a Polynomial
Division Algorithm for Polynomial
If p(x) and g(x) are any two polynomials with g(x) ≠ 0, then we can find polynomials q(x) and r(x) such that
P(x) = g(x) × q(x) + r(x),
where r(x) = 0 or degree of r(x) < degree of g(x).
1. Find the zeroes of the quadratic polynomial 3x2 – 2 and verify the relationship between the zeroes and the coefficients. [2016]
Ans:-
2. Divide the polynomial x4 – 11 x2 + 34x – 12 by x – 2 and find the quotient and the remainder. Also verify the division algorithm. [2016]
Ans:-
3. If one zero of the quadratic polynomial f(x) = 4x2 – 8kx + 8x – 9 is negative of the other, then find the zeroes of kx2 + 3kx + 2. [2015] Ans:- −𝟏 & − 𝟐
4. Obtain all other zeroes of the polynomial x4 + 4x3-2x2-20x -15 if two of its zeroes are √5
and -√5 [2015] Ans:- −𝟏 𝐚𝐧𝐝 − 𝟑
5. If a polynomial x4 – 3x3 – 6x2 + kx – 16 is exactly divisible by x2 – 3x + 2, then find the
value of k.
[2013] Ans:- 24
6. Divide 2x 4– 9x3 + 5x 2 + 3x – 8 bv x 2– 4x+ 1 and verify the division algorithm. [2010]
7. If the polynomial x 4 + 2x 3 + 8x 2 + 12x + 18 is divided by another polynomial x 2 + 5, the remainder comes out to be px + q, find values of p and q. [2009]
Ans:- 𝐩 = 𝟐 & 𝐪 = 𝟑
8. Find all the zeroes of the polynomial 2 x 3 +x 2 – 6 x – 3, if two of its zeroes are -√3 and √3
Ans:- -1/2
9. For what value of p, (-4) is a zero of the polynomial x 2 – 2x – (7p + 3) [2009]
10. Obtain all other zeroes of the polynomial x4 – 3√2x3 – 3x2 + 3√2x – 4, if two of its zeroes Ans :- 3
are √2 and 2√2. Ans:- −1 and 1
11. An NGO decided to distribute books and pencils to the students of a school running by some other NGO. For this, they collected some amount from different number of people. The total amount collected is represented by 4 x4 + 2x3 – 8x2 + 3x – 7. . The amount is equally divided between each of the students. The number of students, who received the amount, is represented by x – 2 + 2x2. After distribution, 5x- 11, amount is left with the NGO which they donated to school for their infrastructure. Find the amount received by each student from the NGO.
What value has been depicted here?
Ans:- (𝟐𝐱𝟐 − 𝟐)
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EX-3.1
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 :
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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 :
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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 :
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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):
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):
Solution :
Question 2(iii):
Solution :
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Question 2(iv):
1,1
Solution :
Question 2(v):
Solution :
Question 2(vi):
4,1
Solution :
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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 :
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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
Solution :
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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 Solutionkey Institute Class 10 Solutions For Maths chapter 3 Polynomials are helpful to complete your math homework.
