CBSE Class 10 Mathematics Polynomials Assignment Set 12

Long Answer Type Questions

Question 1. Verify that \( \frac12, 1, -2 \) are zeroes of the cubic polynomial \( 2x^3 + x^2 - 5x + 2 \). Also verify the relationship between the zeroes and their coefficients.
Answer:
f(1/2) = 2/8 + 1/4 − 5/2 + 2 = 1/4 + 1/4 − 5/2 + 2 = 0; f(1) = 2 + 1 − 5 + 2 = 0; f(−2) = −16 + 4 + 10 + 2 = 0. So the three numbers are zeroes.
Let α = 1/2, β = 1, γ = −2. Sum = 1/2 + 1 − 2 = −1/2 = −(coefficient of x²)/(coefficient of x³) = −1/2. ✓
Sum of products two at a time = (1/2)(1) + (1)(−2) + (−2)(1/2) = 1/2 − 2 − 1 = −5/2 = (coefficient of x)/(coefficient of x³) = −5/2. ✓
Product = (1/2)(1)(−2) = −1 = −(constant term)/(coefficient of x³) = −2/2 = −1. ✓
In simple words: Substitute each number, then compare the sum, pair-sum and product with the coefficients.

Teacher's Note:
1. A cubic has at most three zeroes, so these are all of them.
2. Use −b/a, c/a, −d/a.

Question 2. If α and β are the zeroes of the polynomial \( x^2 + 4x + 3 \), form the polynomial whose zeroes are \( 1 + \frac\beta\alpha \) and \( 1 + \frac\alpha\beta \).
Answer:
α + β = −4 and αβ = 3.
Sum of new zeroes = \( 2 + \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{\alpha^2 + \beta^2 + 2\alpha\beta}{\alpha\beta} = \frac{(\alpha+\beta)^2}{\alpha\beta} = \frac{16}{3} \).
Product = \( \left(1 + \frac\beta\alpha\right)\left(1 + \frac\alpha\beta\right) = 2 + \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{(\alpha+\beta)^2}{\alpha\beta} = \frac{16}{3} \).
Polynomial = \( x^2 - \frac{16}{3}x + \frac{16}{3} \), or 3x² − 16x + 16.
In simple words: Both the sum and the product of the new zeroes simplify to (α + β)²/αβ.

Teacher's Note:
1. Combine into one fraction before substituting.
2. Check: α² + β² = 16 − 6 = 10.

Question 3. Using the factor theorem, factorise \( p(x) = 2x^4 - 7x^3 - 13x^2 + 63x - 45 \).
Answer:
Factors of 45 are ±1, ±3, ±5, ±9, ±15, ±45. Try values: p(1) = 2 − 7 − 13 + 63 − 45 = 0, so (x − 1) is a factor. p(−1) = −112 ≠ 0. p(3) = 162 − 189 − 117 + 189 − 45 = 0, so (x − 3) is a factor.
Dividing by (x − 1)(x − 3) = x² − 4x + 3 gives the quotient 2x² + x − 15.
2x² + x − 15 = 2x² + 6x − 5x − 15 = 2x(x + 3) − 5(x + 3) = (x + 3)(2x − 5).
p(x) = (x − 1)(x − 3)(x + 3)(2x − 5)
In simple words: Find two zeroes by trial, divide, then factorise the quadratic.

Teacher's Note:
1. Trial values come from the factors of the constant term.
2. Check by p(−3) = 0 and p(5/2) = 0.

Question 4. Find the value that must be added to \( f(x) = 3x^4 + 13x^3 - 29x^2 + 100x - 77 \) so that the resulting polynomial is completely divisible by \( g(x) = x^2 + 6x - 7 \).
Answer:
Divide f(x) by g(x) = x² + 6x − 7: the quotient is 3x² − 5x + 22 and the remainder is r(x) = −67x + 77.
Check using g(x) = (x + 7)(x − 1): f(1) = 10 and f(−7) = 546. A remainder mx + n with m + n = 10 and −7m + n = 546 gives m = −67 and n = 77.
f(x) = g(x)q(x) + r(x), so f(x) − r(x) is divisible by g(x). We must add −r(x) = 67x − 77.
Add 67x − 77.
In simple words: Add the negative of the remainder.

Teacher's Note:
1. The remainder has degree less than 2.
2. Use f(1) and f(−7) as a check.

Question 5. Simplify: \( \dfrac{(x-y)^2}{(y-z)(z-x)} + \dfrac{(y-z)^2}{(x-y)(z-x)} + \dfrac{(z-x)^2}{(x-y)(y-z)} \).
Answer:
Put over the common denominator (x − y)(y − z)(z − x): the numerator becomes (x − y)³ + (y − z)³ + (z − x)³.
Since (x − y) + (y − z) + (z − x) = 0, we have u³ + v³ + w³ = 3uvw for u + v + w = 0.
Numerator = 3(x − y)(y − z)(z − x), so the expression = 3.
In simple words: A sum of three cubes with zero sum equals 3 times their product.

Teacher's Note:
1. Recall: if a + b + c = 0 then a³ + b³ + c³ = 3abc.
2. Check x = 0, y = 1, z = 2: the value is 3.

Integer Type Questions

Question 6. Find the value of p for which 2 is one of the zeroes of \( f(x) = x^2 + 3x + p \).
Answer:
f(2) = 4 + 6 + p = 0, so p = −10.
In simple words: Substitute x = 2 and solve for p.

Teacher's Note:
1. A zero makes the polynomial 0.
2. Check: x² + 3x − 10 = (x + 5)(x − 2).

Question 7. Find the value of x in the polynomial \( 2a^2 + 2xa + 5a + 10 \) for which (a + x) is one of its factors.
Answer:
Put a = −x: 2x² − 2x² − 5x + 10 = 0, so −5x + 10 = 0 and x = 2.
Check: 2a² + 4a + 5a + 10 = 2a² + 9a + 10 = (a + 2)(2a + 5).
In simple words: Substitute a = −x in the polynomial.

Teacher's Note:
1. The factor (a + x) vanishes at a = −x.
2. The variable here is a.

Question 8. If the sum of the zeroes of the quadratic polynomial \( 3x^2 - mx + 6 \) is 3, then find the value of m.
Answer:
Sum of zeroes = −(−m)/3 = m/3 = 3, so m = 9.
In simple words: Sum of zeroes = m/3.

Teacher's Note:
1. The question prints k but the polynomial has m.
2. Use −b/a.

Question 9. Find the value of p for which one zero of the polynomial \( 3x^2 - 4x + p \) is the reciprocal of the other.
Answer:
If the zeroes are m and 1/m, their product is 1 = p/3, so p = 3.
In simple words: Reciprocal zeroes multiply to 1.

Teacher's Note:
1. Product of zeroes = c/a = p/3.
2. Check: 3x² − 4x + 3 has D < 0; for real zeroes the question assumes formal zeroes.

Question 10. If (x − 1) is one factor of the quadratic polynomial \( x^2 + 3x + m \), then find the value of m.
Answer:
f(1) = 1 + 3 + m = 0, so m = −4.
In simple words: A factor x − 1 means f(1) = 0.

Teacher's Note:
1. x² + 3x − 4 = (x − 1)(x + 4).
2. Use the factor theorem.

Question 11. If m and n are the zeroes of the polynomial \( x^2 + mx + n \), find the value of m − n.
Answer:
Sum: m + n = −m, so n = −2m. Product: mn = n, so n(m − 1) = 0.
Case n ≠ 0: m = 1 and n = −2, so m − n = 1 + 2 = 3.
(If n = 0 then m = 0 and m − n = 0, the trivial case x² with both zeroes 0. The expected answer is 3.)
In simple words: Use the sum and product of zeroes with the given polynomial.

Teacher's Note:
1. Divide by n only if n ≠ 0.
2. Check: x² + x − 2 has zeroes 1 and −2 = m and n.

Question 12. Find the value of \( a^4 b^3 + a^3 b^4 \), where a and b are the zeroes of \( f(x) = x^2 - 4x + 3 \).
Answer:
a + b = 4 and ab = 3.
a⁴b³ + a³b⁴ = a³b³(a + b) = (ab)³(a + b) = 27 × 4 = 108.
In simple words: Take out a³b³.

Teacher's Note:
1. Zeroes are 1 and 3: 3³·1³·... gives 27 + 81 = 108.
2. Factorise before substituting.

Question 13. Find the value of \( \frac1p + \frac1q + \frac1r \), where p, q and r are the zeroes of the polynomial \( 6x^3 + 3x^2 - 5x + 1 \).
Answer:
pq + qr + rp = −5/6 and pqr = −1/6.
1/p + 1/q + 1/r = (pq + qr + rp)/pqr = (−5/6)/(−1/6) = 5.
In simple words: Combine the fractions over pqr.

Teacher's Note:
1. Use c/a and −d/a.
2. The signs cancel.

Question 14. If the HCF of the polynomials \( x^2 + 2mx - 15 \) and \( (x - 3)(x + n) \) is (x + 5), then find the value of m + n.
Answer:
(x + 5) divides x² + 2mx − 15, so (−5)² − 10m − 15 = 0, giving m = 1 and x² + 2x − 15 = (x − 3)(x + 5).
(x + 5) must also divide (x − 3)(x + n), so (x + n) = (x + 5), and n = 5.
m + n = 1 + 5 = 6.
In simple words: Use the factor theorem for m and the common factor for n.

Teacher's Note:
1. The source says the HCF is (x + 5); both polynomials then also share (x − 3). Read it as: (x + 5) is a common factor.
2. Answer as in the source.

Question 15. Find the degree of the polynomial \( \dfrac{3x^3 + 10x^2 + x - 14}{3x^2 + 13x + 14} \).
Answer:
Divide: 3x³ + 10x² + x − 14 = (3x² + 13x + 14)(x − 1) exactly.
So the expression equals x − 1, which has degree 1.
In simple words: The division is exact, leaving x − 1.

Teacher's Note:
1. Highest power in the numerator minus that in the denominator: 3 − 2 = 1.
2. Check by multiplying back.