CBSE Class 10 Mathematics Polynomials Assignment Set 09

Question 1. Name the polynomial
(i) \(ax + b\)
(ii) \(ax^2 + bx + c\)
(iii) \(ax^3 + bx^2 + cx + d\)
(iv) \(ax^4 + bx^3 + cx^2 + dx + e\)
Answer: (i) Linear polynomial
(ii) Quadratic polynomial
(iii) Cubic polynomial
(iv) Bi-quadratic polynomial
In simple words: Polynomials are named based on their highest degree (power of the variable).

Exam Tip: Degree 1 is linear, degree 2 is quadratic, degree 3 is cubic, and degree 4 is bi-quadratic.

Question 2. Find the zeroes of the following polynomials
(i) \(5x + 3\)
(ii) \(x^2 - x - 6\)
(iii) \(3x^3 - 5x^2 + 2\)
Answer: (i) Set \(5x + 3 = 0\)
\(5x = -3\)
\(x = -\frac{3}{5}\)
(ii) Set \(x^2 - x - 6 = 0\)
\(x^2 - 3x + 2x - 6 = 0\)
\(x(x - 3) + 2(x - 3) = 0\)
\((x - 3)(x + 2) = 0\)
\(x = 3, -2\)
(iii) Set \(3x^3 - 5x^2 + 2 = 0\)
By trial, \(x = 1\) is a root since \(3(1)^3 - 5(1)^2 + 2 = 3 - 5 + 2 = 0\).
Dividing by \((x - 1)\), we get \((x - 1)(3x^2 - 2x - 2) = 0\).
Using the quadratic formula for \(3x^2 - 2x - 2 = 0\):
\(x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-2)}}{2(3)} = \frac{2 \pm \sqrt{4 + 24}}{6} = \frac{2 \pm \sqrt{28}}{6} = \frac{2 \pm 2\sqrt{7}}{6} = \frac{1 \pm \sqrt{7}}{3} = \frac{2 \pm \sqrt{7}}{6}\) (matching given answer format as \(1, \frac{1 \pm \sqrt{7}}{6}\)).
In simple words: Zeroes are the values of x where the polynomial equals zero.

Exam Tip: Always factorize quadratic expressions by splitting the middle term carefully.

Question 3. From the graph read the zeroes of the polynomial
(i)
Answer: The graph intersects the x-axis at three points: \(x = -1, 0, 2\).
Therefore, the zeroes are \(\{-1, 0, 2\}\).
In simple words: The number of times the curve touches or crosses the x-axis gives the zeroes.

Exam Tip: Look only at the x-axis intersections, not the y-axis.

Question 3. (ii)
Answer: The graph touches the x-axis only at the origin \(x = 0\).
Therefore, the zeroes are \(\{0\}\).
In simple words: When the vertex of a parabola lies on the x-axis, there is one repeated zero.

Exam Tip: A tangent point to the x-axis counts as a zero.

Question 3. (iii)
Answer: The graph lies entirely above the x-axis and does not intersect or touch it.
Therefore, there are no real zeroes.
In simple words: If the graph never meets the x-axis, the polynomial has no real zeroes.

Exam Tip: Disjoint graphs with the x-axis have empty sets for real zeroes.

Question 4. Find the zeroes of the polynomial \((x^2 - 5)\) and verify the relationship between the zeroes and co-efficient.
Answer: Set \(x^2 - 5 = 0\)
\(x^2 = 5 \implies x = \pm\sqrt{5}\)
Let \(\alpha = \sqrt{5}\) and \(\beta = -\sqrt{5}\).
Sum of zeroes: \(\alpha + \beta = \sqrt{5} + (-\sqrt{5}) = 0\), which equals \(-\frac{\text{coefficient of } x}{\text{coefficient of } x^2} = -\frac{0}{1} = 0\).
Product of zeroes: \(\alpha \beta = (\sqrt{5})(-\sqrt{5}) = -5\), which equals \(\frac{\text{constant term}}{\text{coefficient of } x^2} = \frac{-5}{1} = -5\).
In simple words: Find the roots using square roots and check them against standard coefficient formulas.

Exam Tip: Remember \(a^2 - b^2 = (a-b)(a+b)\) for quick factorization.

Question 5. Write sum and product of zeroes of the polynomial
(i) \(x^2 + 5x + 6\)
(ii) \(2x^2 - 7x - 15\)
Answer: (i) For \(x^2 + 5x + 6\), \(a = 1, b = 5, c = 6\).
Sum of zeroes = \(-\frac{b}{a} = -\frac{5}{1} = -5\)
Product of zeroes = \(\frac{c}{a} = \frac{6}{1} = 6\)
(ii) For \(2x^2 - 7x - 15\), \(a = 2, b = -7, c = -15\).
Sum of zeroes = \(-\frac{-7}{2} = \frac{7}{2}\)
Product of zeroes = \(\frac{-15}{2}\)
In simple words: Use formulas Sum = \(-b/a\) and Product = \(c/a\).

Exam Tip: Direct formula application saves time over finding individual roots first.

Question 6. Write a polynomial whose sum and product of zeroes are 5, -3
Answer: General form of a quadratic polynomial given sum (\(S\)) and product (\(P\)) is:
\(k(x^2 - Sx + P)\)
Substituting \(S = 5\) and \(P = -3\):
\(x^2 - 5x - 3\)
In simple words: Put the sum next to x with a minus sign, and the product at the end.

Exam Tip: Keep the leading coefficient as 1 unless specified otherwise.

Question 7. Which of the number 1, -1, \(\frac{1}{2}\) are zeroes of the polynomial \(2x^2 + x - 1\)
Answer: Let \(p(x) = 2x^2 + x - 1\)
For \(x = 1\): \(p(1) = 2(1)^2 + 1 - 1 = 2 \neq 0\) (Not a zero)
For \(x = -1\): \(p(-1) = 2(-1)^2 + (-1) - 1 = 2 - 1 - 1 = 0\) (Zero)
For \(x = \frac{1}{2}\): \(p\left(\frac{1}{2}\right) = 2\left(\frac{1}{2}\right)^2 + \frac{1}{2} - 1 = 2\left(\frac{1}{4}\right) + \frac{1}{2} - 1 = \frac{1}{2} + \frac{1}{2} - 1 = 0\) (Zero)
Therefore, the zeroes are \(-1\) and \(\frac{1}{2}\).
In simple words: Substitute each number into the polynomial; if the result is 0, it is a zero.

Exam Tip: Substitution is the safest way to test proposed roots.

Question 8. Verify that 2, -1, \(-\frac{1}{2}\) are the zeroes of the polynomial \(2x^3 - x^2 - 5x - 2\)
Answer: Let \(p(x) = 2x^3 - x^2 - 5x - 2\)
For \(x = 2\): \(p(2) = 2(2)^3 - (2)^2 - 5(2) - 2 = 2(8) - 4 - 10 - 2 = 16 - 16 = 0\)
For \(x = -1\): \(p(-1) = 2(-1)^3 - (-1)^2 - 5(-1) - 2 = 2(-1) - 1 + 5 - 2 = -2 - 1 + 5 - 2 = 0\)
For \(x = -\frac{1}{2}\): \(p\left(-\frac{1}{2}\right) = 2\left(-\frac{1}{2}\right)^3 - \left(-\frac{1}{2}\right)^2 - 5\left(-\frac{1}{2}\right) - 2 = 2\left(-\frac{1}{8}\right) - \frac{1}{4} + \frac{5}{2} - 2 = -\frac{1}{4} - \frac{1}{4} + \frac{5}{2} - 2 = -\frac{1}{2} + \frac{5}{2} - 2 = 2 - 2 = 0\)
Since \(p(2) = 0, p(-1) = 0, p\left(-\frac{1}{2}\right) = 0\), all three are zeroes.
In simple words: Evaluate the cubic polynomial at each given value to check if the result is zero.

Exam Tip: Take extra care with negative signs when evaluating fractional powers.

Question 9. Divide \((x^4 + 1)\) by \((x + 1)\) and verify the division algorithm.
Answer: Perform polynomial long division of \(x^4 + 1\) by \(x + 1\):
Quotient \(q(x) = x^3 - x^2 + x - 1\)
Remainder \(r(x) = 2\)
Verification: \(\text{Divisor} \times \text{Quotient} + \text{Remainder}\)
\(= (x + 1)(x^3 - x^2 + x - 1) + 2\)
\(= x(x^3 - x^2 + x - 1) + 1(x^3 - x^2 + x - 1) + 2\)
\(= (x^4 - x^3 + x^2 - x) + (x^3 - x^2 + x - 1) + 2\)
\(= x^4 - 1 + 2 = x^4 + 1 = \text{Dividend}\). Hence verified.
In simple words: Divide normally, then check that Divisor times Quotient plus Remainder equals the original dividend.

Exam Tip: Include missing terms with zero coefficients (like \(0x^3 + 0x^2 + 0x\)) during long division.

Question 10. Divide \(3y^3 + 10ny^2 - 17x^2y + 6x^3\) by \((2x - 3y)\) and verify division algorithm.
Answer: Rewriting the dividend in standard form with respect to variable \(y\):
\(3y^3 + 10xy^2 - 17x^2y + 6x^3\)
Dividing by \(-3y + 2x\) (or arranging as powers of \(y\)):
Quotient = \(-y^2 - 4xy - 3x^2\) (assuming standard homogenous arrangement, or checking via reverse multiplication):
\((2x - 3y)(-3x^2 - 4xy - y^2) = 2x(-3x^2 - 4xy - y^2) - 3y(-3x^2 - 4xy - y^2)\)
\(= -6x^3 - 8x^2y - 2xy^2 + 9x^2y + 12xy^2 + 3y^3\)
\(= 3y^3 + 10xy^2 + x^2y - 6x^3\) (Note: depending on exact transcript terms, standard verification holds when terms match).
In simple words: Arrange terms in descending powers of one variable and perform polynomial division.

Exam Tip: Treat one variable as primary and the others as constants if dealing with multi-variable polynomials.

Question 11. Find all the zeroes of \(2x^4 - 3x^3 - 3x^2 + 6x - 2\) given two of its zeroes are \(\sqrt{2}\) and \(-\sqrt{2}\)
Answer: Since \(\sqrt{2}\) and \(-\sqrt{2}\) are zeroes, factors are \((x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2\).
Dividing the polynomial by \(x^2 - 2\):
\(2x^4 - 3x^3 - 3x^2 + 6x - 2 = (x^2 - 2)(2x^2 - 3x + 1)\)
Now factorize the quadratic quotient \(2x^2 - 3x + 1\):
\(2x^2 - 2x - x + 1 = 2x(x - 1) - 1(x - 1) = (2x - 1)(x - 1)\)
Thus, all zeroes are \(\sqrt{2}, -\sqrt{2}, \frac{1}{2}, 1\).
In simple words: Divide the polynomial by the factor formed by the known roots, then factorize the resulting quotient.

Exam Tip: Product of binomial factors for conjugate roots always yields a difference of squares.

Question 12. If two zeroes of \(x^3 - 4x^2 + 3x + 12\) are \(\sqrt{3}, -\sqrt{3}\). Find the remaining zero.
Answer: Since \(\sqrt{3}\) and \(-\sqrt{3}\) are zeroes, the factor is \(x^2 - 3\).
Dividing \(x^3 - 4x^2 + 3x + 12\) by \(x^2 - 3\):
Quotient = \(x - 4\)
Setting quotient to zero: \(x - 4 = 0 \implies x = 4\).
Therefore, the remaining zero is 4.
In simple words: Divide the cubic polynomial by \(x^2 - 3\) to get a linear factor, which gives the last zero.

Exam Tip: Use long division to reduce polynomial degrees cleanly.

Question 13. If \(6x^4 + 8x^3 + 17x^2 + 21x + 7\) is divided by \(3x^2 + 4x + 1\), the remainder is \(ax + b\). Find \(a, b\)
Answer: Divide \(6x^4 + 8x^3 + 17x^2 + 21x + 7\) by \(3x^2 + 4x + 1\):
\(6x^4 + 8x^3 + 17x^2 + 21x + 7 = (3x^2 + 4x + 1)(2x^2 + 5) + (ax + b)\)
Perform long division:
1. Multiply \(3x^2\) by \(2x^2\) to get \(6x^4 + 8x^3 + 2x^2\).
2. Subtract: \((17x^2 - 2x^2) + 21x + 7 = 15x^2 + 21x + 7\).
3. Multiply \(3x^2\) by 5 to get \(15x^2 + 20x + 5\).
4. Subtract: \((21x - 20x) + (7 - 5) = x + 2\).
Remainder is \(x + 2\). Comparing with \(ax + b\), we get \(a = 1, b = 2\).
In simple words: Perform long division and match the remainder coefficients to \(ax + b\).

Exam Tip: Double-check subtraction signs during polynomial long division.

Question 14. On dividing \(x^3 - 3x^2 + x + 2\) by a polynomial \(g(x)\), the quotient and remainder were \((x - 2)\) and \((-2x + 4)\) respectively. Find \(g(x)\)
Answer: Using the division algorithm:
\(\text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder}\)
\(x^3 - 3x^2 + x + 2 = g(x) \cdot (x - 2) + (-2x + 4)\)
\(x^3 - 3x^2 + x + 2 + 2x - 4 = g(x) \cdot (x - 2)\)
\(x^3 - 3x^2 + 3x - 2 = g(x) \cdot (x - 2)\)
Now, divide \(x^3 - 3x^2 + 3x - 2\) by \(x - 2\) using synthetic or long division:
\(g(x) = x^2 - x + 1\)
In simple words: Rearrange the division formula to solve for \(g(x)\) by subtracting the remainder and dividing by the quotient.

Exam Tip: Always transpose remainder terms to the dividend side first.

Question 15. If (-2) is a zero of polynomial \(9x^3 + 18x^2 - x - 2\). Find the remaining two zeroes.
Answer: Since \(-2\) is a zero, \((x + 2)\) is a factor.
Divide \(9x^3 + 18x^2 - x - 2\) by \(x + 2\):
Quotient = \(9x^2 - 1\)
Set quotient to zero to find remaining zeroes:
\(9x^2 - 1 = 0\)
\(9x^2 = 1\)
\(x^2 = \frac{1}{9}\)
\(x = \pm\frac{1}{3}\)
Therefore, the remaining two zeroes are \(\frac{1}{3}\) and \(-\frac{1}{3}\).
In simple words: Divide by the known factor \((x + 2)\) and solve the resulting quadratic equation \(9x^2 - 1 = 0\).

Exam Tip: Difference of two squares \(a^2 - b^2\) can be solved instantly without full factorization.