CBSE - Polynomials

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points

(i) The number of zeroes is 0 as the graph does not cut the x-axis at any point.

(ii) The number of zeroes is 1 as the graph intersects the x-axis at only 1 point.

(iii) The number of zeroes is 3 as the graph intersects the x-axis at 3 points.

(iv) The number of zeroes is 2 as the graph intersects the x-axis at 2 points.

(v) The number of zeroes is 4 as the graph intersects the x-axis at 4 points.

(vi) The number of zeroes is 3 as the graph intersects the x-axis at 3 points

(i) x² – 2x – 8
= (x - 4) (x + 2)
The value of x² – 2x – 8 is zero when x - 4 = 0 or x + 2 = 0, i.e., when x = 4 or x = -2
Therefore, the zeroes of x² – 2x – 8 are 4 and -2.

Sum of zeroes = 4 + (-2) = 2 = -(-2)/1 = -(Coefficient of x)/Coefficient of x²

Product of zeroes = 4 × (-2) = -8 = -8/1 = Constant term/Coefficient of x²

(ii) 4s² – 4s + 1
= (2s-1)²
The value of 4s² - 4s + 1 is zero when 2s - 1 = 0, i.e., s = 1/2

Therefore, the zeroes of 4s² - 4s + 1 are 1/2 and 1/2.

Sum of zeroes = 1/2 + 1/2 = 1 = -(-4)/4 = -(Coefficient of s)/Coefficient of s²
Product of zeroes = 1/2 × 1/2 = 1/4 = Constant term/Coefficient of s².

(iii) 6x² – 3 – 7x
= 6x² – 7x – 3
= (3x + 1) (2x - 3)
The value of 6x² - 3 - 7x is zero when 3x + 1 = 0 or 2x - 3 = 0, i.e., x = -1/3 or x = 3/2

Therefore, the zeroes of 6x² - 3 - 7x are -1/3 and 3/2.

Sum of zeroes = -1/3 + 3/2 = 7/6 = -(-7)/6 = -(Coefficient of x)/Coefficient of x²
Product of zeroes = -1/3 × 3/2 = -1/2 = -3/6 = Constant term/Coefficient of x².

(iv) 4u² + 8u
= 4u² + 8u + 0
= 4u(u + 2)
The value of 4u² + 8u is zero when 4u = 0 or u + 2 = 0, i.e., u = 0 or u = - 2
Therefore, the zeroes of 4u² + 8u are 0 and - 2.

Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of u)/Coefficient of u²
Product of zeroes = 0 × (-2) = 0 = 0/4 = Constant term/Coefficient of u².

(v) t² – 15
= t² - 0.t - 15
= (t - √15) (t + √15)
The value of t² - 15 is zero when t - √15 = 0 or t + √15 = 0, i.e., when t = √15 or t = -√15
Therefore, the zeroes of t² - 15 are √15 and -√15.Sum of zeroes = √15 + -√15 = 0 = -0/1 = -(Coefficient of t)/Coefficient of t²
Product of zeroes = (√15) (-√15) = -15 = -15/1 = Constant term/Coefficient of u².

(vi) 3x² – x – 4
= (3x - 4) (x + 1)
The value of 3x² – x – 4 is zero when 3x - 4 = 0 and x + 1 = 0,i.e., when x = 4/3 or x = -1
Therefore, the zeroes of 3x² – x – 4 are 4/3 and -1.
 

Sum of zeroes = 4/3 + (-1) = 1/3 = -(-1)/3 = -(Coefficient of x)/Coefficient of x²

Product of zeroes = 4/3 × (-1) = -4/3 = Constant term/Coefficient of x².

(i) 1/4 , -1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 1/4 = -b/a
αß = -1 = -4/4 = c/a
If a = 4, then b = -1, c = -4
Therefore, the quadratic polynomial is 4x² - x -4.

(ii) √2 , 1/3
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = √2 = 3√2/3 = -b/a
αß = 1/3 = c/a
If a = 3, then b = -3√2, c = 1
Therefore, the quadratic polynomial is 3x² -3√2x +1.

(iii) 0, √5
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 0 = 0/1 = -b/a

αß = √5 = √5/1 = c/a
If a = 1, then b = 0, c = √5
Therefore, the quadratic polynomial is x² + √5.

(iv) 1, 1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 1 = 1/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -1, c = 1
Therefore, the quadratic polynomial is x² - x +1.

(v) -1/4 ,1/4
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = -1/4 = -b/a
αß = 1/4 = c/a
If a = 4, then b = 1, c = 1
Therefore, the quadratic polynomial is 4x² + x +1.

(vi) 4,1
Let the polynomial be ax² + bx + c, and its zeroes be α and ß
α + ß = 4 = 4/1 = -b/a
αß = 1 = 1/1 = c/a
If a = 1, then b = -4, c = 1
Therefore, the quadratic polynomial is x² - 4x +1.

(i) Let us assume the division of 6x² + 2x + 2 by 2
Here, p(x) = 6x² + 2x + 2
g(x) = 2
q(x) = 3x² + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x² + 2x + 2 = 2x (3x² + x + 1)
Hence, division algorithm is satisfied.

(ii) Let us assume the division of x³+ x by x²,
Here, p(x) = x³ + x
g(x) = x²
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + x = (x² ) × x + x
x³ + x = x³ + x
Thus, the division algorithm is satisfied.

(iii) Let us assume the division of x³+ 1 by x².
Here, p(x) = x³ + 1
g(x) = x²
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + 1 = (x² ) × x + 1
x³ + 1 = x³ + 1
Thus, the division algorithm is satisfied.

2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x⁴ – 6x³ – 26x² + 138x – 35.
Let x = 2±√3
So, x-2 = ±√3
On squaring, we get x² - 4x + 4 = 3,
⇒ x² - 4x + 1= 0

Now, dividing p(x) by x² - 4x + 1

∴ p(x) = x⁴ - 6x³ - 26x² + 138x - 35
= (x² - 4x + 1) (x2 - 2x - 35)
= (x² - 4x + 1) (x2 - 7x + 5x - 35)
= (x² - 4x + 1) [x(x - 7) + 5 (x - 7)]
= (x² - 4x + 1) (x + 5) (x - 7)

∴ (x + 5) and (x - 7) are other factors of p(x).
∴ - 5 and 7 are other zeroes of the given polynomial.

On dividing x⁴ – 6x³ + 16x² – 25x + 10 by x² – 2x + k

∴ Remainder = (2k - 9)x - (8 - k)k + 10 

But the remainder is given as x+ a. 

On comparing their coefficients,

2k - 9 = 1

⇒ k = 10 

⇒ k = 5 and,

-(8-k)k + 10 = a

⇒ a = -(8 - 5)5 + 10 =- 15 + 10 = -5 

Hence, k = 5 and a = -5