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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
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
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
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
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