147-NCERT-10-2-Polynomials - Ex-2.4

NCERT

10th Mathematics

Exercise 2.4

Topic: 2 Polynomials

Click here for ⇨ NCERT-10-2-Polynomials - Ex-2.3

EXERCISE 2.4

1. Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also, verify the relationship between the zeroes and the coefficients in each case:

(i) 2x³ + x² – 5x + 2;     1/2, 1, -2

(ii) x³ – 4x² + 5x – 2;     2, 1, 1

Explanation:

1) We know that if 𝝰, 𝛃, and 𝜸 are the zeroes of the cubic polynomial

ax³ + bx² + cx + d, then 

𝝰 + 𝛃 + 𝜸 = - b/a = [- (coeficient of x²)/(coeficient of x³)]

𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 = c/a = [(coeficient of x)/(coeficient of x³)]
𝝰 x 𝛃 x 𝜸 = - d/a = [- (constant)/(coeficient of x³)]

2)  If 𝝰, 𝛃, and 𝜸 are the zeroes of the cubic polynomial

p(x) = ax³ + bx² + cx + d, then p(𝝰) = p(𝛃) = p(𝜸) = 0.

Solution:

(i) 2x³ + x² – 5x + 2;     1/2, 1, -2

1) Let 𝝰 = 1/2, 𝛃 = 1, and 𝜸 = -2

2) Let p(x) = 2x³ + x² – 5x + 2

p(1/2) = 2(1/2)³ + (1/2)² – 5(1/2) + 2

= 2(1/8) + (1/4) – 5(1/2) + 2 

= (1/4) + (1/4) – (5/2) + 2

= [(1 + 1 - 10 + 8)/4]

= [(0)/4]

= 0 ---------------- this shows that 1/2 is the zeroes of p(x). 

p(1) = 2(1)³ + (1)² – 5(1) + 2

= 2(1) + (1) – 5(1) + 2 

= (2) + (1) – (5) + 2

= [(2 + 1 - 5 + 2)]

= 0 ---------------- this shows that 1 is the zeroes of p(x).

p(-2) = 2(-2)³ + (-2)² – 5(-2) + 2

= 2(-8) + (4) – 5(-2) + 2 

= (-16) + (4) – (-10) + 2

= [(-16 + 4 + 10 + 2)]

= [(-16 + 16)]

= 0 ---------------- this shows that -2 is the zeroes of p(x).

3) So 1/2, 1, -2 are the zeroes of the cubic polynomial p(x) = 2x³ + x² – 5x + 2.

4) Here 𝝰 = 1/2, 𝛃 = 1, and 𝜸 = -2 and a = 2, b = 1, c = - 5, and d = 2.

5) Now we will verify the relation between zeroes and the coefficients.

a) 𝝰 + 𝛃 + 𝜸 = - b/a

LHS = 𝝰 + 𝛃 + 𝜸 

= 1/2 + 1 - 2

= - 1/2 -------------------- equation 1

RHS = - b/a 

= - 1/2

= - 1/2 -------------------- equation 2

From equation 1 and equation 2, LHS = RHS.

b) 𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 = c/a

LHS = 𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 

= [(1/2) x (1)] + [(1) x (- 2)] + [(- 2) x (1/2)]

= [(1/2)] + [(- 2)] + [(- 1)]

= (1/2) - 3 

= - 5/2 -------------------- equation 3

RHS = c/a 

= (- 5)/2

= - 5/2 -------------------- equation 4

From equation 3 and equation 4, LHS = RHS. 

c) 𝝰 x 𝛃 x 𝜸 = - d/a

LHS = 𝝰 x 𝛃 x 𝜸 

= (1/2) x (1) x (- 2)

= - 1 -------------------- equation 5

RHS = - d/a 

= - 2/2

= - 1 -------------------- equation 6

From equation 5 and equation 6, LHS = RHS. 

6) So, the relation between zeroes and coefficient is verified.

 

(ii) x³ – 4x² + 5x – 2;     2, 1, 1

1) Let 𝝰 = 2, 𝛃 = 1, and 𝜸 = 1

2) Let p(x) = x³ – 4x² + 5x – 2

p(2) = (2)³ – 4(2)² + 5(2) – 2

= (8) – 4(4) + 5(2) – 2 

= 8 – 16 + 10 – 2

= 18 - 18

= 0 ---------------- this shows that 2 is the zeroes of p(x). 

p(1) = (1)³ – 4(1)² + 5(1) – 2

= (1) – 4(1) + 5(1) – 2 

= 1 – 4 + 5 – 2

= 6 – 6

= 0 ---------------- this shows that 1 is the zeroes of p(x).

p(1) = (1)³ – 4(1)² + 5(1) – 2

= (1) – 4(1) + 5(1) – 2 

= 1 – 4 + 5 – 2

= 6 – 6

= 0 ---------------- this shows that 1 is the zeroes of p(x).

3) So 2, 1, 1 are the zeroes of the cubic polynomial p(x) = x³ – 4x² + 5x – 2.

4) Here 𝝰 = 2, 𝛃 = 1, and 𝜸 = 1 and a = 1, b = - 4, c = 5, and d = - 2.

5) Now we will verify the relation between zeroes and the coefficients.

a) 𝝰 + 𝛃 + 𝜸 = - b/a

LHS = 𝝰 + 𝛃 + 𝜸 

= 2 + 1 + 1

= 4 -------------------- equation 1

RHS = - (- 4)/1 

= 4 -------------------- equation 2

From equation 1 and equation 2, LHS = RHS.

b) 𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 = c/a

LHS = 𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 

= [(2) x (1)] + [(1) x (1)] + [(1) x (2)]

= [(2)] + [(1)] + [(2)]

= 2 + 1 + 2 

= 5 -------------------- equation 3

RHS = c/a 

= (5)/1

= 5 -------------------- equation 4

From equation 3 and equation 4, LHS = RHS. 

c) 𝝰 x 𝛃 x 𝜸 = - d/a

LHS = 𝝰 x 𝛃 x 𝜸 

= (2) x (1) x (1)

= 2 -------------------- equation 5

RHS = - d/a 

= - (- 2)/1

= 2 -------------------- equation 6

From equation 5 and equation 6, LHS = RHS.

6) So, the relation between zeroes and coefficient is verified.

Q2. Find a cubic polynomial with the sum, the sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively. 

Explanation:

1) We know that if 𝝰, 𝛃, and 𝜸 are the zeroes of the cubic polynomial 

ax³ + bx² + cx + d, then 

𝝰 + 𝛃 + 𝜸 = - b/a = [- (coeficient of x²)/(coeficient of x³)]

𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 = c/a = [(coeficient of x)/(coeficient of x³)]
𝝰 x 𝛃 x 𝜸 = - d/a = [- (constant)/(coeficient of x³)]

2)  If 𝝰, 𝛃, and 𝜸 are the zeroes of the cubic polynomial

p(x) = ax³ + bx² + cx + d, then p(𝝰) = p(𝛃) = p(𝜸) = 0.

Solution:

1) Let ax³ + bx² + cx + d be our cubic polynomial.

2) Here 𝝰 + 𝛃 + 𝜸 = 2, 𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 = - 7, 𝝰 x 𝛃 x 𝜸 = - 14.

3) We know that 

𝝰 + 𝛃 + 𝜸 = - b/a

2/1 = - b/a ------------- therefore a = 1, and b = -2.

 

𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 = c/a

-7/1 = c/a ------------- therefore a = 1, and c = - 7.

 

𝝰 x 𝛃 x 𝜸 = - d/a

-14/1 = -d/a ------------- therefore a = 1, and c = 14.

4) our cubic polynomial will be x³ - 2x² - 7x + 14. 

Q3. If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.

Explanation:

1) We know that if 𝝰, 𝛃, and 𝜸 are the zeroes of the cubic polynomial 

px³ + qx² + rx + s, then 

𝝰 + 𝛃 + 𝜸 = - q/p = [- (coeficient of x²)/(coeficient of x³)]

𝝰𝛃 + 𝛃𝜸 + 𝜸𝝰 = r/p = [(coeficient of x)/(coeficient of x³)]
𝝰 x 𝛃 x 𝜸 = - s/p = [- (constant)/(coeficient of x³)]

2)  If 𝝰, 𝛃, and 𝜸 are the zeroes of the cubic polynomial

f(x) = px³ + qx² + rx + s, then f(𝝰) = f(𝛃) = f(𝜸) = 0.

Solution:

1) (a – b), (a), (a + b) are zeroes of polynomial x³ – 3x² + x + 1.

2) So, 𝝰 = ( a - b), 𝛃 = a, and 𝜸 = ( a + b), p = 1, q = - 3, r = 1, s = 1.

3) As 𝝰 + 𝛃 + 𝜸 = - q/p

(a – b) + (a) + (a + b) = - (- 3)/1

(3a) = 3

So, a = 1 --------------- equation 1.

4) As 𝝰 x 𝛃 x 𝜸 = - s/p

(a – b) x (a) x (a + b) = - (1)/1

(a² – b²) x a = - 1 --------------- equation 2.

From equation 1, put a = 1 in equation 2, we get,

(1² – b²) x 1 = - 1

1 – b² = - 1

b² = 2

So, b =  ± √2.

5) So, a = 1 and b =  ± √2.

Q4. If two zeroes of the polynomial x⁴ – 6x³ – 26x² + 138x – 35 are 2 ± √3, find other zeroes.

Explanation:

1) Let p(x) = x⁴ – 6x³ – 26x² + 138x – 35 and 2 - √3 and 2 + √3 are zeroes of p(x).

2) So, [x - (2 - √3)] x [x + (2 + √3)] is a factor of x⁴ – 6x³ – 26x² + 138x – 35.

3) Now we will find other factors of p(x).

Solution:

1) Here [x - (2 - √3)] x [x + (2 + √3)] is a factor of x⁴ – 6x³ – 26x² + 138x – 35.

2) [x - (2 - √3)] x [x + (2 + √3)] = [(x - 2) - √3] x [(x + 2) + √3].

= (x - 2)² - (√3)²

= (x² - 4x + 4) - (3)

= x² - 4x + 1 

3) So, (x² - 4x + 1) divides the polynomial x⁴ – 6x³ – 26x² + 138x – 35.

4) Here (x² – 4x – 1) is one factor of x⁴ – 6x³ – 26x² + 138x – 35 and the other one

is (x² – 2x – 35). 

5) Now we will find all factors of (x² – 2x – 35).

 x² – 2x – 35 = x² – 7x + 5x – 35

= (x² – 7x) + (5x – 35)

= x(x – 7) + 5(x – 7)

= (x – 7) x (x + 5)  

6) Here x = (2 - √3), x =(2 + √3), x = - 5, and x = 7 are the other zeroes of x⁴ – 6x³ – 26x² + 138x – 35.

7) All zeroes of x⁴ – 6x³ – 26x² + 138x – 35 are (2 - √3), (2 + √3), x = - 5, and x = 7.

Q5. If the polynomial x⁴ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.

Explanation:

1) We know that Dividend = Divisor × Quotient + Remainder

2) 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).

Solution:

1) Here p(x) = x⁴ – 6x³ + 16x² – 25x + 10 and g(x) = x² – 2x + k.

2) Now we will find the quotient and remainder using long division.

3) Here remainder r(x) is given as r(x) = (x + a) ---------------------equation 1

4) From equation 1, we have r(x) = (x + a),

so, (-9+2k)x + (10-8k+k²) = (x + a)  ------------- equation 2

5) From equation 2, we have, 

     (-9 + 2k) = 1

2k = 10

k = 5  ------------- equation 3

6) From equation 2, we have,

10 - 8k + k² = a from equation 3, put k = 5, we get,

10 - 8(5) + (5)² = a

10 - 40 + 25 = a

- 30 + 25 = a 

a = - 5

7) So, k = 5 and a = - 5.

Click here for ⇨ NCERT-10-3-Pair of Linear Equations in Two Variables - Ex-3.1

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