At Saralstudy, we are providing you with the solution of Class 10 Mathematics Polynomials according to the latest NCERT (CBSE) Book guidelines prepared by expert teachers. Here we are trying to give you a detailed answer to the questions of the entire topic of this chapter so that you can get more marks in your examinations by preparing the answers based on this lesson. We are trying our best to give you detailed answers to all the questions of all the topics of Class 10 Mathematics Polynomials so that you can prepare for the exam according to your own pace and your speed.
Total no of zeroes of a polynomial equation = the number of times the curve intersect x-axis
(i) x2 – 2x – 8
= x – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x + 2) (x – 4)
The value of x2 – 2x – 8 is zero if (x + 2) = 0 and (x – 4) = 0
x = -2 or x = 4
Sum of zeroes = (-2 + 4) = 2 = - coefficient of x
coefficient of x2
Product of zeroes = (-2) × 4 = -8 = Constant term
coefficient of x2
(ii) 4s2 – 4s + 1
= 4s2 – 2s – 2s + 1
= 2s (2s – 1) – 1 (2s – 1)
= ( 2s – 1 ) ( 2s – 1 )
The value of 4s2 – 4s + 1 is zero , if (2s-1) = 0 and (2s-1 ) = 0
s = 1/2 , 1/2
Sum of zeroes = (1/2 + 1/2) = 1 - coefficient of x
coefficient of x2
Product of zeroes =1/2 × 1/2 = 1/4 = constant term
coefficient of x2
(iii) 6x2 –7x – 3
= 6x – 9x + 2x – 3
= 3x (2x – 3) + 1(2x – 3)
= (3x + 1) (2x – 3)
The value of 6x2 –7x – 3 is zero, if (3x + 1) = 0 and (2x – 3) = 0
X = -1 /3 , 3/2
Sum of zeroes = ( -1/3 + 3/2) = 7/6 = - coefficient of x
coefficient of x2
Product of zeroes = -1/3 × 3/2 = -3/2 = constant term
coefficient of x2
(iv) 4u2+8u
4u(u+2)
The value of 4u2+8u is zero, if 4u = 0 and (u+2) =0
u = 0, - 2
Sum of zeroes = ( 0+ (-2)) = -2 = - coefficient of x
coefficient of x2
Product of zeroes = (-2) × 0 = 0 = constant term
coefficient of x2
(v)
(vi)
3x2–x–4
3x – 4x + 3x – 4
= x (3x – 4) + 1 (3x – 4)
The value of 3x – x + 4 is zero, if (3x – 4) = 0 and (x + 1) = 0
Sum of zeroes = [4/3 + ( -1)] = 1/3 = - coefficient of x
coefficient of x2
Product of zeroes = (-1) × 4/3 = -4/3 = constant term
coefficient of x2
(i) Let α, β are the zeroes of the polynomial ax2+ bx +c, therefore
Sum of zeroes (α + β) = 1/4 = -b/a
Product of zeroes (αβ) = -1 = c/a
On comparing,
a = 4, b = -1 and c = - 4
Hence, the required polynomial is 4x2 –x – 4
(ii)
(iii)
Let α, β are the zeroes of the polynomial ax2+ bx +c, therefore
Sum of zeroes (α + β) = 0 = -b/a
Product of zeroes (αβ) = √5 = √5/1 = c/a
On comparing,
a = 1, b = 0 and c = √5
Hence, the required polynomial is x2 + √5
(i) Given ,
Dividend = p(x) = x2 – 3x2 + 5x – 3
Divisor = g(x) = x2 – 2
Quotient = x – 3
Reminder = 7x – 9
(ii) Given,
Dividend = p(x) = x4 – 3x2 + 4x + 5
Divisor = g(x) = x2 + 1 – x
Exercise 3 ( Page No. : 36 )
Note: If on dividing the second polynomial by first we get zero remainder then we say that first Is factor of second polynomial.
(i) Given ,
First polynomial = t2-3
Second polynomial = 2t4 +3t3-2t2 -9t-12
As we can see the remainder is 0. Thereofre we can say that first polynomial is a factor of second polynomial.
(ii) Given,
First polynomial = x2+3x+1
Second polynomial = 3x4 + 5x3 – 7x2 + 2x + 2
Exercise 3 ( Page No. : 36 )
Exercise 3 ( Page No. : 36 )
Given ,
Dividend p(x) = x3 - 3x2 + x + 2
Quotient = x – 2
Remainder = -2x + 4
Let divisor = g(x)
As we know ,
Dividend = Divisor × Quotient + Remainder
x3- 3x2 +x+ 2 = g(x) × (x – 2) + (-2x + 4)
x 3- 3x2 + x + 2 – (-2x + 4) = g(x) × (x – 2 )
Therefore, g(x) × (x – 2 ) = x3 – x2 + x + 2
Now we will divide p(x) by quotient x – 2 to find divisor g(x)
Therefore , g(x) = ( x2 – x + 1 )
we know that
(i) Deg P(x) = deg g (x)
The degree of dividend or quotient can be equal, only if the divisor is a constant (degree 0)
Then, let p(x) = 3x2 – 6x + 5
Let g(x) = 3
Therefore, q(x) = x2 – 2x +1 and r(x) = 2
(ii) Deg q(x) = deg r(x)
Let p(x) = x2 + 1
Let g(x) = x
Therefore, q(x) = x + 1 and r(x) = 0
Here, we can see the degree of quotient is equal to the degree of remainder.
Hence, division algorithm is satisfied here.
(iii) deg r(x) = 0
The degree of remainder is zero, only if the remainder left after division algorithm is Constant.
Let p(x) = x2 + 1
Let g(x) = x
Therefore, q(x)= x and r(x) = 1
Here we can see the degree of remainder is zero.
Hence division algorithm is satisfied here.
(i) Here f(x) = 2x³ + x² - 5x + 2
Given roots of f(x) are ½, 1, -2
F(1/2) = 2×(1/2)³ + (1/2)² - 5(1/2 ) + 2 = 0
F(1) = 2(1)³ + 1² - 5(1) + 2 = 0
F(-2) = 2(-2)³ + (-2)² - 5(-2) + 2 = 0
Hence, ½, 1 and -2 are the zeroes of f(x).
Therefore, sum of zeroes = -b/a -1/2
Sum of product of zeroes taken two at a time = c/a = -5/2
Product of zeroes = -d/a = 2
(ii) Let the f(x) = ax³ + bx² + c + d
Let α, β and γ be the zeroes of the polynomial f(x).
Then, sum of zeroes = -b/a = 2/1 ………………(i)
Sum of product of zeroes taken two at a time = c/a = -7. ………………..(ii)
Product of zeroes = -d/a = -14 ……………….(iii)
From equation (i), (ii) and (iii) we have
a = 1 , b = -2 , c = -7 and d = 14
Therefore the required polynomial on putting the values of a, b, c and d
F(x) = x³ - 2x² - 7x + 14
Let the p(x) = ax3+bx2+ cx + d
Sum of zeroes and α, β and γ be the zeroes.
Then, α, β and γ = -b/ a = 2/1 …………………..(i)
αβ + βγ + γα= c/a = -7 …………………….(ii)
‘ αβγ = -d /a = -14 ……………………………..(iii)
From equation (i), (ii) and (iii), we get
a = 1, b = -2, c = -7 and d = 14
Therefore, the required polynomial on putting the value of a, b, c and d is P(x) = x3 - 2x2 – 7x + 14
Given, p(x) = x3 - 3x2 + x + 1
And zeroes are given as a – b, a, a + b
Now, comparing the given polynomial with general expression, we get;
∴ ax3+bx2+ cx + d = x3 – 3x2+ x + 1
a = 1, b = -3, c = 1 and d = 1
Sum of zeroes = a – b + a + a + b
-b/a = 3a
Putting the values b and a
- (-3)/1 = 3a
a = 1
Thus, the zeroes are 1 - b, 1, 1 + b.
Now, product of zeroes = 1(1 – b) (1 + b)
d/a = 1 - b2
-1/1 = 1- b2
b2 = 1 + 1 = 2
b = √2
Hence, 1, -√2, 1, 1 + √2 are the zeroes of x3 – 3x2 + x + 1
Exercise 4 ( Page No. : 37 )
Given,
Divisor = x2 – 2x + k
Dividend = x4 – 6x3 + 16x2 – 25x + 10
Remainder = x + a
As we know that,
Dividend = divisor quotient + remainder
x4 – 6x3 + 16x2 – 25x + 10 = x2 – 2x + k quotient + (x + a)
x4 – 6x3 + 16x2 – 25x + 10 – (x + a) = x2 – 2x + k quotient
x4 – 6x3 + 16x2 – 26x + 10 – a = quotient
x2 – 2x + k
If the polynomial x4 – 6x3 + 16x2 – 26x + 10 – a is divided by x2 – 2x + k remainder comes out to be zero.
Therefore, By equating the remainder with zero, we have
(-10 + 2k) = 0 => 2k = 10 => k = 5
Or, 10 – a – 8k + k2 = 0
Putting the value of k, we get
10 – a – 8(5) + (5)2 = 0
10 – a – 40 + 25 = 0
- a – 5 = 0 => a = -5
Hence, k = 5 and a = -5
