CBSE Class 10 Mathematics Polynomials Assignment Set 11

Very Short Answer Type Questions

Question 1. Form the quadratic polynomial whose zeroes are 4 and 6.
Answer:
Sum = 4 + 6 = 10 and product = 4 × 6 = 24.
Polynomial = x² − (sum)x + product = x² − 10x + 24.
In simple words: Use x² − (sum)x + (product).

Teacher's Note:
1. Any non-zero multiple k(x² − 10x + 24) also works.
2. Check: (x − 4)(x − 6) = x² − 10x + 24.

Question 2. Form the quadratic polynomial whose sum and product of zeroes are 6 and 7.
Answer:
Polynomial = x² − 6x + 7.
In simple words: Put the given sum and product in the standard form.

Teacher's Note:
1. The sign of the middle term is minus.
2. Zeroes are 3 ± √2.

Question 3. If α, β and γ are the zeroes of the polynomial \( x^3 - 3x^2 + x + 1 \), find the value of α + β + γ.
Answer:
For \( ax^3 + bx^2 + cx + d \), α + β + γ = −b/a = −(−3)/1 = 3.
In simple words: Sum of zeroes is −(coefficient of x²) / (coefficient of x³).

Teacher's Note:
1. Read a = 1, b = −3.
2. Mind the sign.

Question 4. If x = 2 and x = 0 are the zeroes of the polynomial \( f(x) = 2x^3 - 5x^2 + ax + b \), then find the value of a + b.
Answer:
f(0) = b = 0.
f(2) = 16 − 20 + 2a + b = 0, so 2a = 4 and a = 2.
a + b = 2 + 0 = 2.
In simple words: Use f(0) = 0 and f(2) = 0.

Teacher's Note:
1. Substitute each zero in turn.
2. f(0) gives b directly.

Question 5. Find the remainder when \( f(x) = x^3 - 6x^2 + 2x - 4 \) is divided by g(x) = 1 − 2x.
Answer:
g(x) = 0 at x = 1/2, so the remainder is \( f\left(\frac12\right) = \frac18 - \frac64 + 1 - 4 = -\frac{35}{8} \).
Remainder = \( -\frac{35}{8} \).
In simple words: Put x = 1/2 in f(x).

Teacher's Note:
1. The zero of 1 − 2x is 1/2.
2. Use a common denominator of 8.

Question 6. If x = 3 is a root of \( f(x) = x^3 - 4x^2 + 6k \), then find the value of k.
Answer:
f(3) = 27 − 36 + 6k = 0, so 6k = 9 and k = 3/2.
In simple words: Substitute x = 3 and solve for k.

Teacher's Note:
1. A root makes the polynomial zero.
2. 27 − 36 = −9.

Question 7. Find the remainder when \( f(x) = x^3 + x^2 + 2x + 3 \) is divided by x − 1.
Answer:
Remainder = f(1) = 1 + 1 + 2 + 3 = 7.
In simple words: By the remainder theorem put x = 1.

Teacher's Note:
1. The zero of x − 1 is 1.
2. No long division needed.

Question 8. Find the remainder when the polynomial \( P(x) = x^4 + 6x^3 + 13x^2 + 12x + 6 \) is divided by g(x) = x + 2.
Answer:
Remainder = P(−2) = 16 − 48 + 52 − 24 + 6 = 2.
In simple words: Put x = −2.

Teacher's Note:
1. The zero of x + 2 is −2.
2. Take care with the signs of the odd powers.

Question 9. Find the value of a for which (x − 3) is a factor of \( f(x) = x^2 - ax + 9 \).
Answer:
f(3) = 9 − 3a + 9 = 0, so 3a = 18 and a = 6.
In simple words: A factor means f(3) = 0.

Teacher's Note:
1. Apply the factor theorem.
2. Check: x² − 6x + 9 = (x − 3)².

Question 10. Show that (x + a) is always a factor of \( x^n + a^n \) when n is an odd positive integer.
Answer:
Let \( f(x) = x^n + a^n \). By the factor theorem (x + a) is a factor if f(−a) = 0.
\( f(-a) = (-a)^n + a^n = -a^n + a^n = 0 \) since n is odd.
Hence (x + a) is a factor of xⁿ + aⁿ for odd n.
In simple words: An odd power keeps the minus sign, so the two terms cancel.

Teacher's Note:
1. For even n, (−a)ⁿ = aⁿ and the sum is 2aⁿ, not zero.
2. Use the factor theorem with x = −a.

Short Answer Type Questions

Question 11. Find the zeroes of the quadratic polynomial \( 6x^2 - 13x + 6 \) and then verify the relation between the zeroes and its coefficients.
Answer:
\( 6x^2 - 13x + 6 = 6x^2 - 4x - 9x + 6 = 2x(3x - 2) - 3(3x - 2) = (3x - 2)(2x - 3) \).
Zeroes: \( \frac23 \) and \( \frac32 \).
Sum = \( \frac23 + \frac32 = \frac{13}{6} = -\frac{-13}{6} \). Product = \( \frac23 \times \frac32 = 1 = \frac66 \).
Both relations are verified.
In simple words: Split the middle term, find the zeroes, then check sum and product.

Teacher's Note:
1. Find two numbers with product 36 and sum −13: −4 and −9.
2. Compare with −b/a and c/a.

Question 12. Verify the relation between the zeroes and the coefficients of the polynomial \( ax^2 + bx + c \), a ≠ 0.
Answer:
Let α and β be the zeroes. By the factor theorem \( ax^2 + bx + c = k(x - \alpha)(x - \beta) = kx^2 - k(\alpha+\beta)x + k\alpha\beta \).
Comparing coefficients: a = k, b = −k(α + β), c = kαβ.
So \( \alpha + \beta = -\frac ba \) and \( \alpha\beta = \frac ca \). That is, sum of zeroes = −(coefficient of x)/(coefficient of x²) and product = (constant term)/(coefficient of x²).
In simple words: Write the polynomial as k(x − α)(x − β) and compare.

Teacher's Note:
1. The constant k equals a.
2. This is the derivation of the standard results.

Question 13. If α and β are the zeroes of the polynomial \( ax^2 + bx + c \), then find the value of (i) α − β (ii) α² + β².
Answer:
α + β = −b/a, αβ = c/a.
(i) \( (\alpha-\beta)^2 = (\alpha+\beta)^2 - 4\alpha\beta = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2} \), so \( \alpha - \beta = \pm\frac{\sqrt{b^2 - 4ac}}{a} \).
(ii) \( \alpha^2 + \beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = \frac{b^2}{a^2} - \frac{2c}{a} = \) \( \frac{b^2 - 2ac}{a^2} \).
In simple words: Use (α ± β)² identities with the sum and product.

Teacher's Note:
1. The sign of α − β depends on which zero is larger.
2. Keep a² in the denominator.

Question 14. If α and β are the zeroes of the quadratic polynomial \( ax^2 - bx + c \), then find the value of α² − β².
Answer:
Here α + β = b/a and αβ = c/a.
\( \alpha^2 - \beta^2 = (\alpha+\beta)(\alpha-\beta) \), and \( (\alpha-\beta)^2 = \frac{b^2}{a^2} - \frac{4c}{a} = \frac{b^2 - 4ac}{a^2} \).
So \( \alpha^2 - \beta^2 = \) \( \pm\frac{b\sqrt{b^2 - 4ac}}{a^2} \) (+ if α > β, − if α < β).
In simple words: Factor as (α + β)(α − β) and use the sum and product.

Teacher's Note:
1. The source answer omits the ± sign.
2. Note the middle term is −bx, so the sum is +b/a.

Question 15. Find a cubic polynomial whose zeroes are 1, 2 and −3.
Answer:
Sum = 1 + 2 − 3 = 0. Sum of products two at a time = 1·2 + 2·(−3) + (−3)·1 = −7. Product = 1 × 2 × (−3) = −6.
Polynomial = x³ − (sum)x² + (sum of products)x − product = x³ − 0x² − 7x + 6.
x³ − 7x + 6
In simple words: Use the sum, pair-sum and product of the zeroes.

Teacher's Note:
1. Check x = 1: 1 − 7 + 6 = 0.
2. Any non-zero multiple also works.

Question 16. If α and β are the zeroes of the polynomial \( ax^2 + bx + c \), then form the polynomial whose zeroes are \( \frac1\alpha \) and \( \frac1\beta \).
Answer:
Sum of new zeroes = \( \frac{\alpha+\beta}{\alpha\beta} = \frac{-b/a}{c/a} = -\frac bc \). Product = \( \frac{1}{\alpha\beta} = \frac ac \).
Polynomial = \( x^2 + \frac bc x + \frac ac \), or cx² + bx + a.
In simple words: Reciprocals reverse the order of the coefficients.

Teacher's Note:
1. Check with x² − 3x + 2 (zeroes 1, 2): reciprocals give 2x² − 3x + 1.
2. Assume c ≠ 0.

Question 17. Divide \( p(x) = 3x^3 + 4x + 11 \) by \( g(x) = x^2 - 3x + 2 \) to find the quotient and remainder.
Answer:
Write p(x) = 3x³ + 0x² + 4x + 11. First term 3x: subtract 3x³ − 9x² + 6x, leaving 9x² − 2x + 11. Next term 9: subtract 9x² − 27x + 18, leaving 25x − 7.
Since the degree of 25x − 7 is less than the degree of g(x), we stop.
Quotient = 3x + 9, remainder = 25x − 7.
Check: (3x + 9)(x² − 3x + 2) + 25x − 7 = 3x³ + 4x + 11.
In simple words: Long division, writing a 0x² placeholder.

Teacher's Note:
1. Arrange both polynomials in descending powers.
2. Verify using dividend = divisor × quotient + remainder.

Question 18. On dividing \( 3x^3 + 4x^2 + 5x - 13 \) by a polynomial g(x), the quotient and remainder were 3x + 10 and 16x − 43, respectively. Find g(x).
Answer:
Dividend − remainder = divisor × quotient: 3x³ + 4x² + 5x − 13 − (16x − 43) = 3x³ + 4x² − 11x + 30.
g(x) = (3x³ + 4x² − 11x + 30) ÷ (3x + 10) = x² − 2x + 3 (long division leaves remainder 0).
g(x) = x² − 2x + 3
In simple words: Remove the remainder, then divide by the quotient.

Teacher's Note:
1. Check: (3x + 10)(x² − 2x + 3) = 3x³ + 4x² − 11x + 30.
2. Use p = g × q + r.

Question 19. For a given polynomial \( p(x) = x^4 - 6x^3 - 26x^2 + 138x - 35 \), the sum of two zeroes is 4. Find the sum of the other two zeroes.
Answer:
For a quartic ax⁴ + bx³ + …, the sum of all four zeroes is −b/a = −(−6)/1 = 6.
Sum of the other two zeroes = 6 − 4 = 2.
In simple words: The four zeroes add up to 6.

Teacher's Note:
1. Read b = −6, a = 1.
2. Check: the zeroes are 7, −5, 2 ± √3.

Question 20. Find the sum of the real values of y satisfying \( x^2 + x^2y^2 + x^2y^4 = 525 \) and \( x + xy + xy^2 = 35 \).
Answer:
Divide the squared second equation by the first: \( \frac{x^2(1+y+y^2)^2}{x^2(1+y^2+y^4)} = \frac{1225}{525} = \frac73 \).
Since \( 1 + y^2 + y^4 = (1 + y + y^2)(1 - y + y^2) \), this gives \( \frac{1 + y + y^2}{1 - y + y^2} = \frac73 \).
3 + 3y + 3y² = 7 − 7y + 7y², so 4y² − 10y + 4 = 0, that is 2y² − 5y + 2 = 0 and y = 2 or y = 1/2.
Both give real x (y = 2: x = 5; y = 1/2: x = 20, and both satisfy the first equation). Sum of the real values of y = 2 + 1/2 = 5/2.
In simple words: Divide the squared second equation by the first to get an equation in y alone.

Teacher's Note:
1. Factorise 1 + y² + y⁴.
2. Check x = 5, y = 2: 25 + 100 + 400 = 525.