Exercise 2.1 1 Question
The graphs of y = p(x) are given below for some polynomials p(x). Find the number of zeroes of p(x) in each case.
(i) Parabola opening upward –
does not intersect x-axis
Zeroes: 0
(ii) Parabola – touches x-axis at
one point
Zeroes: 1
(iii) Parabola – cuts x-axis at
two points
Zeroes: 2
(iv) Parabola opening downward –
cuts x-axis at two points
Zeroes: 2
(v) Curve – cuts x-axis at
one point
Zeroes: 1
(vi) Curve – cuts x-axis at
three points
Zeroes: 3
Exercise 2.2 2 Questions
Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.
(i) x² – 2x – 8
Solution:
x² – 2x – 8 = x² – 4x + 2x – 8
= x(x – 4) + 2(x – 4)
= (x – 4)(x + 2)
∴ Zeroes are: x = 4 and x = –2
Verification:
For ax² + bx + c, we know: Sum of zeroes = –b/a, Product = c/a
Here: a = 1, b = –2, c = –8
Sum of zeroes = 4 + (–2) = 2
–b/a = –(–2)/1 = 2 âœ"
Product of zeroes = 4 × (–2) = –8
c/a = –8/1 = –8 âœ"
(ii) 4s² – 4s + 1
Solution:
4s² – 4s + 1 = (2s – 1)²
∴ Zeroes are: s = ½ and s = ½ (both equal)
Verification:
a = 4, b = –4, c = 1
Sum of zeroes = ½ + ½ = 1
–b/a = –(–4)/4 = 1 âœ"
Product of zeroes = ½ × ½ = ¼
c/a = 1/4 = ¼ âœ"
(iii) 6x² – 3 – 7x
Solution:
Rearrange: 6x² – 7x – 3
= 6x² – 9x + 2x – 3
= 3x(2x – 3) + 1(2x – 3)
= (2x – 3)(3x + 1)
∴ Zeroes are: x = 3/2 and x = –1/3
Verification:
a = 6, b = –7, c = –3
Sum of zeroes = 3/2 + (–1/3) = 7/6
–b/a = –(–7)/6 = 7/6 âœ"
Product of zeroes = (3/2) × (–1/3) = –1/2
c/a = –3/6 = –1/2 âœ"
(iv) 4u² + 8u
Solution:
4u² + 8u = 4u(u + 2)
∴ Zeroes are: u = 0 and u = –2
Verification:
a = 4, b = 8, c = 0
Sum of zeroes = 0 + (–2) = –2
–b/a = –8/4 = –2 âœ"
Product of zeroes = 0 × (–2) = 0
c/a = 0/4 = 0 âœ"
(v) t² – 15
Solution:
t² – 15 = (t – √15)(t + √15)
∴ Zeroes are: t = √15 and t = –√15
Verification:
a = 1, b = 0, c = –15
Sum of zeroes = √15 + (–√15) = 0
–b/a = –0/1 = 0 âœ"
Product of zeroes = (√15)(–√15) = –15
c/a = –15/1 = –15 âœ"
(vi) 3x² – x – 4
Solution:
3x² – x – 4 = 3x² – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4)
= (3x – 4)(x + 1)
∴ Zeroes are: x = 4/3 and x = –1
Verification:
a = 3, b = –1, c = –4
Sum of zeroes = 4/3 + (–1) = 1/3
–b/a = –(–1)/3 = 1/3 âœ"
Product of zeroes = (4/3) × (–1) = –4/3
c/a = –4/3 = –4/3 âœ"
Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
(i) 1/4, –1
Let the quadratic polynomial be ax² + bx + c, and its zeroes be α and β.
Given: α + β = 1/4, αβ = –1
If a = 1, then –b = 1/4 â‡' b = –1/4, and c = –1
One such polynomial: x² – (1/4)x – 1
Multiply by 4: 4x² – x – 4 is also a valid answer.
(ii) √2, 1/3
Given: α + β = √2, αβ = 1/3
The polynomial is: k[x² – (sum)x + product], k ≠0
For k = 1: x² – √2 x + 1/3
Multiply by 3: 3x² – 3√2 x + 1
(iii) 0, √5
Given: α + β = 0, αβ = √5
The polynomial is: x² – 0·x + √5 = x² + √5
(iv) 1, 1
Given: α + β = 1, αβ = 1
The polynomial is: x² – x + 1
(v) –1/4, 1/4
Given: α + β = –1/4, αβ = 1/4
The polynomial is: x² + (1/4)x + 1/4
Multiply by 4: 4x² + x + 1
(vi) 4, 1
Given: α + β = 4, αβ = 1
The polynomial is: x² – 4x + 1
Exercise 2.3 5 Questions
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
(i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
Solution using Division Algorithm:
Divide: x³ ÷ x² = x (first term of quotient)
Multiply: x(x² – 2) = x³ – 2x
Subtract: (x³ – 3x² + 5x – 3) – (x³ – 2x) = –3x² + 7x – 3
Divide: –3x² ÷ x² = –3 (next term)
Multiply: –3(x² – 2) = –3x² + 6
Subtract: (–3x² + 7x – 3) – (–3x² + 6) = 7x – 9
∴ Quotient = x – 3 and Remainder = 7x – 9
Verify: p(x) = g(x) × q(x) + r(x)
(x² – 2)(x – 3) + (7x – 9) = x³ – 3x² – 2x + 6 + 7x – 9 = x³ – 3x² + 5x – 3 âœ"
(ii) p(x) = xâ´ – 3x² + 4x + 5, g(x) = x² + 1 – x
Rearrange g(x): g(x) = x² – x + 1
Divide: xⴠ÷ x² = x²
Multiply: x²(x² – x + 1) = xâ´ – x³ + x²
Subtract: (xâ´ + 0x³ – 3x² + 4x + 5) – (xâ´ – x³ + x²) = x³ – 4x² + 4x + 5
Divide: x³ ÷ x² = x
Multiply: x(x² – x + 1) = x³ – x² + x
Subtract: (x³ – 4x² + 4x + 5) – (x³ – x² + x) = –3x² + 3x + 5
Divide: –3x² ÷ x² = –3
Multiply: –3(x² – x + 1) = –3x² + 3x – 3
Subtract: (–3x² + 3x + 5) – (–3x² + 3x – 3) = 8
∴ Quotient = x² + x – 3 and Remainder = 8
(iii) p(x) = xâ´ – 5x + 6, g(x) = 2 – x²
Rearrange g(x): g(x) = –x² + 2. For convenience, use g(x) = –x² + 2.
Divide: xⴠ÷ (–x²) = –x²
Multiply: –x²(–x² + 2) = xâ´ – 2x²
Subtract: (xâ´ + 0x² – 5x + 6) – (xâ´ – 2x²) = 2x² – 5x + 6
Divide: 2x² ÷ (–x²) = –2
Multiply: –2(–x² + 2) = 2x² – 4
Subtract: (2x² – 5x + 6) – (2x² – 4) = –5x + 10
∴ Quotient = –x² – 2 and Remainder = –5x + 10
Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
(i) t² – 3, 2tâ´ + 3t³ – 2t² – 9t – 12
Divide: 2tⴠ÷ t² = 2t²
Multiply: 2t²(t² – 3) = 2tâ´ – 6t²
Subtract: (2tâ´ + 3t³ – 2t² – 9t – 12) – (2tâ´ – 6t²) = 3t³ + 4t² – 9t – 12
Divide: 3t³ ÷ t² = 3t
Multiply: 3t(t² – 3) = 3t³ – 9t
Subtract: (3t³ + 4t² – 9t – 12) – (3t³ – 9t) = 4t² – 12
Divide: 4t² ÷ t² = 4
Multiply: 4(t² – 3) = 4t² – 12
Subtract: (4t² – 12) – (4t² – 12) = 0
Since remainder = 0, t² – 3 is a factor of 2tâ´ + 3t³ – 2t² – 9t – 12. âœ"
(ii) x² + 3x + 1, 3xâ´ + 5x³ – 7x² + 2x + 2
Divide: 3xⴠ÷ x² = 3x²
Multiply: 3x²(x² + 3x + 1) = 3xⴠ+ 9x³ + 3x²
Subtract: (3xâ´ + 5x³ – 7x² + 2x + 2) – (3xâ´ + 9x³ + 3x²) = –4x³ – 10x² + 2x + 2
Divide: –4x³ ÷ x² = –4x
Multiply: –4x(x² + 3x + 1) = –4x³ – 12x² – 4x
Subtract: (–4x³ – 10x² + 2x + 2) – (–4x³ – 12x² – 4x) = 2x² + 6x + 2
Divide: 2x² ÷ x² = 2
Multiply: 2(x² + 3x + 1) = 2x² + 6x + 2
Subtract: (2x² + 6x + 2) – (2x² + 6x + 2) = 0
Since remainder = 0, x² + 3x + 1 is a factor âœ"
(iii) x³ – 3x + 1, xâµ – 4x³ + x² + 3x + 1
Divide: xⵠ÷ x³ = x²
Multiply: x²(x³ – 3x + 1) = xâµ – 3x³ + x²
Subtract: (xâµ – 4x³ + x² + 3x + 1) – (xâµ – 3x³ + x²) = –x³ + 3x + 1
Divide: –x³ ÷ x³ = –1
Multiply: –1(x³ – 3x + 1) = –x³ + 3x – 1
Subtract: (–x³ + 3x + 1) – (–x³ + 3x – 1) = 2
Remainder = 2 (not zero), so x³ – 3x + 1 is NOT a factor. ✗
Obtain all other zeroes of 3xâ´ + 6x³ – 2x² – 10x – 5, if two of its zeroes are √(5/3) and –√(5/3).
Given zeroes: α = √(5/3) and β = –√(5/3)
Product: α × β = –5/3. Sum: α + β = 0.
The quadratic factor corresponding to these zeroes: x² – (sum)x + product = x² – 5/3
Multiply by 3: 3x² – 5 is a factor of the polynomial.
Divide 3xâ´ + 6x³ – 2x² – 10x – 5 by (3x² – 5):
3xⴠ÷ 3x² = x²
Multiply: x²(3x² – 5) = 3xâ´ – 5x²
Subtract: (3xâ´ + 6x³ – 2x² – 10x – 5) – (3xâ´ – 5x²) = 6x³ + 3x² – 10x – 5
6x³ ÷ 3x² = 2x
Multiply: 2x(3x² – 5) = 6x³ – 10x
Subtract: (6x³ + 3x² – 10x – 5) – (6x³ – 10x) = 3x² – 5
3x² ÷ 3x² = 1
Multiply: 1(3x² – 5) = 3x² – 5
Subtract: (3x² – 5) – (3x² – 5) = 0
∴ Quotient = x² + 2x + 1 = (x + 1)²
The remaining zeroes come from (x + 1)² = 0 â‡' x = –1 and x = –1
∴ All zeroes are: √(5/3), –√(5/3), –1, –1
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).
Given: p(x) = x³ – 3x² + x + 2, q(x) = x – 2, r(x) = –2x + 4
Using the Division Algorithm: p(x) = g(x) × q(x) + r(x)
â‡' g(x) × q(x) = p(x) – r(x)
p(x) – r(x) = (x³ – 3x² + x + 2) – (–2x + 4)
= x³ – 3x² + x + 2 + 2x – 4
= x³ – 3x² + 3x – 2
Now: g(x) = (x³ – 3x² + 3x – 2) ÷ (x – 2)
Divide: x³ ÷ x = x²
Multiply: x²(x – 2) = x³ – 2x²
Subtract: (x³ – 3x² + 3x – 2) – (x³ – 2x²) = –x² + 3x – 2
Divide: –x² ÷ x = –x
Multiply: –x(x – 2) = –x² + 2x
Subtract: (–x² + 3x – 2) – (–x² + 2x) = x – 2
Divide: x ÷ x = 1
Multiply: 1(x – 2) = x – 2
Subtract: (x – 2) – (x – 2) = 0
∴ g(x) = x² – x + 1
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and:
(i) deg p(x) = deg q(x)
Let: p(x) = 2x² + 3x + 1, g(x) = 2
Then q(x) = x² + (3/2)x + (1/2) with remainder r(x) = 0
Here deg p(x) = deg q(x) = 2 âœ"
(ii) deg q(x) = deg r(x)
Let: p(x) = x³ + x² + x + 1, g(x) = x² – 1
Then q(x) = x + 1 (degree 1), remainder r(x) = 2x + 2 (degree 1)
Here deg q(x) = deg r(x) = 1 âœ"
(iii) deg r(x) = 0
Let: p(x) = x² + 2x + 3, g(x) = x
Then q(x) = x + 2, remainder r(x) = 3 (degree 0)
Here deg r(x) = 0 âœ"
Exercise 2.4 5 Questions – Optional
Verify that the numbers given alongside 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; ½, 1, –2
Verifying zeroes:
p(½) = 2(1/8) + 1/4 – 5(1/2) + 2 = 1/4 + 1/4 – 5/2 + 2 = 0 âœ"
p(1) = 2(1)³ + 1² – 5(1) + 2 = 2 + 1 – 5 + 2 = 0 âœ"
p(–2) = 2(–8) + 4 – 5(–2) + 2 = –16 + 4 + 10 + 2 = 0 âœ"
Relationship with coefficients (ax³ + bx² + cx + d):
α + β + γ = –b/a = –1/2
½ + 1 + (–2) = –½ = –1/2 âœ"
αβ + βγ + γα = c/a = –5/2
(½)(1) + (1)(–2) + (–2)(½) = ½ – 2 – 1 = –5/2 = –5/2 âœ"
αβγ = –d/a = –2/2 = –1
½ × 1 × (–2) = –1 âœ"
(ii) x³ – 4x² + 5x – 2; 2, 1, 1
Verifying zeroes:
p(2) = 8 – 16 + 10 – 2 = 0 âœ"
p(1) = 1 – 4 + 5 – 2 = 0 âœ"
Relationship verification:
a = 1, b = –4, c = 5, d = –2
Sum: 2 + 1 + 1 = 4 = –(–4)/1 = 4 âœ"
Pair products: (2)(1)+(1)(1)+(1)(2) = 2+1+2 = 5 = 5/1 âœ"
Product: 2 × 1 × 1 = 2 = –(–2)/1 = 2 âœ"
Find a cubic polynomial with the sum of zeroes, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, –7, –14 respectively.
For a cubic polynomial ax³ + bx² + cx + d (a ≠0):
Sum of zeroes = –b/a = 2
Sum of pairwise products = c/a = –7
Product of zeroes = –d/a = –14
For a = 1: b = –2, c = –7, d = 14
∴ The polynomial is: x³ – 2x² – 7x + 14
Factor check: x³ – 2x² – 7x + 14 = (x – 2)(x² – 7) = (x – 2)(x – √7)(x + √7)
Zeroes: 2, √7, –√7 â†' Sum = 2 âœ", pair sum = 0 – 7 = –7 âœ", product = –14 âœ"
If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b.
Given polynomial: x³ – 3x² + x + 1, so: a₃ = 1, aâ‚‚ = –3, aâ‚ = 1, aâ‚€ = 1
Sum of zeroes: (a – b) + a + (a + b) = 3a = –aâ‚‚/a₃ = 3
â‡' a = 1
Sum of products taken two at a time:
(a – b)a + a(a + b) + (a – b)(a + b) = aâ‚/a₃ = 1
Put a = 1: (1 – b)(1) + (1)(1 + b) + (1 – b)(1 + b) = 1
1 – b + 1 + b + (1 – b²) = 1
2 + 1 – b² = 1 â‡' b² = 2 â‡' b = ±√2
∴ a = 1, b = ±√2
If two zeroes of the polynomial xâ´ – 6x³ – 26x² + 138x – 35 are 2 ± √3, find the other zeroes.
Given: α = 2 + √3, β = 2 – √3
Sum: α + β = 4. Product: αβ = (2)² – 3 = 1
The quadratic factor: x² – 4x + 1
Divide xâ´ – 6x³ – 26x² + 138x – 35 by x² – 4x + 1:
xⴠ÷ x² = x²
x²(x² – 4x + 1) = xâ´ – 4x³ + x²
Subtract: (xâ´ – 6x³ – 26x² + 138x – 35) – (xâ´ – 4x³ + x²) = –2x³ – 27x² + 138x – 35
–2x³ ÷ x² = –2x
–2x(x² – 4x + 1) = –2x³ + 8x² – 2x
Subtract: (–2x³ – 27x² + 138x – 35) – (–2x³ + 8x² – 2x) = –35x² + 140x – 35
–35x² ÷ x² = –35
–35(x² – 4x + 1) = –35x² + 140x – 35
Subtract: (–35x² + 140x – 35) – (–35x² + 140x – 35) = 0
Quotient = x² – 2x – 35 = (x – 7)(x + 5)
∴ The other zeroes are: 7 and –5
If the polynomial xâ´ – 6x³ + 16x² – 25x + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a. Find k and a.
Let p(x) = xâ´ – 6x³ + 16x² – 25x + 10 and g(x) = x² – 2x + k.
Given: r(x) = x + a
By division algorithm: p(x) = g(x) × q(x) + r(x)
Let q(x) = x² + bx + c. Then:
(x² – 2x + k)(x² + bx + c) + (x + a) = xâ´ – 6x³ + 16x² – 25x + 10
Expand: (x² – 2x + k)(x² + bx + c) = xâ´ + (b – 2)x³ + (c – 2b + k)x² + (kc – 2c + kb)x + kc
Adding r(x): ... + x + a
Compare coefficients with xâ´ – 6x³ + 16x² – 25x + 10:
xâ´: 1 = 1 âœ"
x³: b – 2 = –6 â‡' b = –4
x²: c – 2b + k = c + 8 + k = 16 â‡' c + k = 8 ... (1)
x¹: kc – 2c + kb + 1 = kc – 2c + k(–4) + 1 = –25 â‡' kc – 2c – 4k = –26 ... (2)
xâ°: kc + a = 10 ... (3)
From (1): c = 8 – k
Substitute in (2): k(8 – k) – 2(8 – k) – 4k = –26
8k – k² – 16 + 2k – 4k = –26
–k² + 6k – 16 = –26 â‡' k² – 6k – 10 = 0
(k – 5)(k – 2) = 0, but... let's recheck.
–k² + 6k = –10 â‡' k² – 6k = 10 â‡' k² – 6k – 10 = 0
k = 5 or k = 2. Testing: For k = 5, c = 3. From (3): 5×3 + a = 10 â‡' a = –5
∴ k = 5 and a = –5
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