CBSE Class 10 Mathematics Polynomials Assignment Set 03

Exam Tip: Always equate the remainder obtained from polynomial long division or the remainder theorem to zero when a polynomial is stated to be exactly divisible.

Question 2. If two zeros of the polynomial \(f(x) = x^4 - 6x^3 - 26x^2 + 138x - 35\) are \(2 \pm \sqrt{3}\). Find the other zeros.
Answer: Let the given zeros be \(x = 2 + \sqrt{3}\) and \(x = 2 - \sqrt{3}\).
Sum of zeros = \((2 + \sqrt{3}) + (2 - \sqrt{3}) = 4\)
Product of zeros = \((2 + \sqrt{3})(2 - \sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1\)
The quadratic factor corresponding to these zeros is \(x^2 - (\text{sum})x + \text{product} = x^2 - 4x + 1\).
Now, divide the given polynomial \(f(x)\) by \(x^2 - 4x + 1\):
\(x^4 - 6x^3 - 26x^2 + 138x - 35 = (x^2 - 4x + 1)(x^2 - 2x - 35)\)
Factorizing the quotient quadratic polynomial \(x^2 - 2x - 35\):
\(x^2 - 7x + 5x - 35 = 0\)
\(x(x - 7) + 5(x - 7) = 0\)
\((x - 7)(x + 5) = 0\)
Thus, the other two zeros are 7 and -5.
In simple words: Combine the two given roots into a quadratic factor, divide the big polynomial by it, and then factorize the resulting quotient to find the remaining roots.

Exam Tip: When irrational zeros like \(a \pm \sqrt{b}\) are given, form the quadratic factor using \(x^2 - (2a)x + (a^2 - b) = 0\).

Question 3. Find the Quadratic polynomial whose sum and product of zeros are \(\sqrt{2} + 1, \frac{1}{\sqrt{2} + 1}\).
Answer: Sum of zeros \((S) = \sqrt{2} + 1\)
Product of zeros \((P) = \frac{1}{\sqrt{2} + 1} = \frac{\sqrt{2} - 1}{(\sqrt{2} + 1)(\sqrt{2} - 1)} = \frac{\sqrt{2} - 1}{2 - 1} = \sqrt{2} - 1\)
The required quadratic polynomial is given by:
\(P(x) = x^2 - Sx + P\)
\(P(x) = x^2 - (\sqrt{2} + 1)x + (\sqrt{2} - 1)\)
In simple words: Rationalize the product of zeros first, then substitute both sum and product into the standard quadratic polynomial formula.

Exam Tip: Always simplify denominators containing square roots by rationalization before substituting them into coefficient formulas.

Question 4. If \(\alpha, \beta\) are the zeros of the polynomial \(2x^2 - 4x + 5\) find the value of (a) \(\alpha^2 + \beta^2\) (b) \((\alpha - \beta)^2\).
Answer: Given polynomial \(p(x) = 2x^2 - 4x + 5\).
Sum of zeros: \(\alpha + \beta = -\frac{-4}{2} = 2\)
Product of zeros: \(\alpha\beta = \frac{5}{2}\)
(a) \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\)
\(\alpha^2 + \beta^2 = (2)^2 - 2\left(\frac{5}{2}\right) = 4 - 5 = -1\)
(b) \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\)
\((\alpha - \beta)^2 = (2)^2 - 4\left(\frac{5}{2}\right) = 4 - 10 = -6\)
In simple words: Use algebraic identities to express \(\alpha^2 + \beta^2\) and \((\alpha - \beta)^2\) in terms of sum \((\alpha + \beta)\) and product \((\alpha\beta)\).

Exam Tip: Memorize standard algebraic identity transformations: \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\).

Question 5. If \(\alpha, \beta\) are the zeros of the polynomial \(x^2 + 8x + 6\) frame a Quadratic polynomial whose zeros are (a) \(\frac{1}{\alpha}\) and \(\frac{1}{\beta}\) (b) \(1 + \frac{\beta}{\alpha}\) and \(1 + \frac{\alpha}{\beta}\).
Answer: For polynomial \(x^2 + 8x + 6\), sum \(\alpha + \beta = -8\) and product \(\alpha\beta = 6\).
(a) Let new zeros be \(S_1 = \frac{1}{\alpha}\) and \(S_2 = \frac{1}{\beta}\).
Sum of new zeros = \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{-8}{6} = -\frac{4}{3}\)
Product of new zeros = \(\left(\frac{1}{\alpha}\right)\left(\frac{1}{\beta}\right) = \frac{1}{\alpha\beta} = \frac{1}{6}\)
Required quadratic polynomial = \(x^2 - \left(-\frac{4}{3}\right)x + \frac{1}{6} = x^2 + \frac{4}{3}x + \frac{1}{6}\)
(b) Let new zeros be \(1 + \frac{\beta}{\alpha}\) and \(1 + \frac{\alpha}{\beta}\).
Sum = \(\left(1 + \frac{\beta}{\alpha}\right) + \left(1 + \frac{\alpha}{\beta}\right) = 2 + \left(\frac{\alpha^2 + \beta^2}{\alpha\beta}\right)\)
\(= 2 + \frac{(\alpha + \beta)^2 - 2\alpha\beta}{\alpha\beta} = 2 + \frac{(-8)^2 - 2(6)}{6} = 2 + \frac{64 - 12}{6} = 2 + \frac{52}{6} = 2 + \frac{26}{3} = \frac{32}{3}\)
Product = \(\left(1 + \frac{\beta}{\alpha}\right)\left(1 + \frac{\alpha}{\beta}\right) = 1 + \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 1 = 2 + \frac{\alpha^2 + \beta^2}{\alpha\beta} = \frac{32}{3}\)
Required quadratic polynomial = \(x^2 - \frac{32}{3}x + \frac{32}{3}\)
In simple words: Find the sum and product of the new roots by converting them into expressions of \((\alpha + \beta)\) and \((\alpha\beta)\).

Exam Tip: For symmetric expressions of roots, always substitute \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\).

Question 6. On dividing the polynomial \(4x^4 - 5x^3 - 39x^2 - 46x - 2\) by the polynomial \(g(x)\) the quotient is \(x^2 - 3x - 5\) and the remainder is \(-5x + 8\). Find the polynomial \(g(x)\).
Answer: By division algorithm: \(p(x) = g(x) \cdot q(x) + r(x)\)
\(g(x) = \frac{p(x) - r(x)}{q(x)}\)
Given:
\(p(x) = 4x^4 - 5x^3 - 39x^2 - 46x - 2\)
\(q(x) = x^2 - 3x - 5\)
\(r(x) = -5x + 8\)
Now, \(p(x) - r(x) = 4x^4 - 5x^3 - 39x^2 - 46x - 2 - (-5x + 8)\)
\(= 4x^4 - 5x^3 - 39x^2 - 41x - 10\)
Dividing \(p(x) - r(x)\) by \(q(x)\) using long division:
\(\frac{4x^4 - 5x^3 - 39x^2 - 41x - 10}{x^2 - 3x - 5} = 4x^2 + 7x + 2\)
Therefore, \(g(x) = 4x^2 + 7x + 2\).
In simple words: Subtract the remainder from the dividend polynomial, then divide the result by the quotient polynomial to get \(g(x)\).

Exam Tip: Ensure signs are carefully distributed when subtracting the remainder: \(p(x) - (-5x + 8) = p(x) + 5x - 8\).

Question 7. If the squared difference of the zeros of the quadratic polynomial \(x^2 + px + 45\) is equal to 144, find the value of \(p\).
Answer: Let the zeros be \(\alpha\) and \(\beta\).
For \(p(x) = x^2 + px + 45\):
\(\alpha + \beta = -p\)
\(\alpha\beta = 45\)
Given condition: \((\alpha - \beta)^2 = 144\)
We know that \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta\)
\((-p)^2 - 4(45) = 144\)
\(p^2 - 180 = 144\)
\(p^2 = 144 + 180 = 324\)
\(p = \pm\sqrt{324} = \pm18\)
In simple words: Express the squared difference of roots in terms of sum and product, substitute the coefficients, and solve for \(p\).

Exam Tip: Do not forget both positive and negative roots when taking square roots: \(p^2 = 324 \implies p = \pm 18\).

Question 8. If \(\alpha, \beta\) are the zeros of a Quadratic polynomial such that \(\alpha + \beta = 24\), \(\alpha - \beta = 8\). Find a Quadratic polynomial having \(\alpha\) and \(\beta\) as its zeros.
Answer: Given system of equations:
\(\alpha + \beta = 24\) --- (1)
\(\alpha - \beta = 8\) --- (2)
Adding (1) and (2):
\(2\alpha = 32 \implies \alpha = 16\)
Substitute \(\alpha = 16\) in (1):
\(16 + \beta = 24 \implies \beta = 8\)
Sum of zeros = \(\alpha + \beta = 24\)
Product of zeros = \(\alpha\beta = 16 \times 8 = 128\)
Required quadratic polynomial = \(k(x^2 - (\text{sum})x + \text{product}) = k(x^2 - 24x + 128)\), where \(k\) is a real constant.
In simple words: Solve the simultaneous linear equations to find individual roots \(\alpha\) and \(\beta\), calculate their product, and write the polynomial.

Exam Tip: Always include the constant factor \(k\) when asked to find "a" quadratic polynomial given its zeros.

Question 9. If \(\alpha, \beta\) are the zeroes of the polynomial \(2x^2 - 4x + 5\), then find the value of (a) \(\alpha^2 + \beta^2\) (b) \(\frac{1}{\alpha} + \frac{1}{\beta}\) (c) \((\alpha - \beta)^2\) (d) \(\frac{1}{\alpha^2} + \frac{1}{\beta^2}\) (e) \(\alpha^3 + \beta^3\).
Answer: For \(p(x) = 2x^2 - 4x + 5\):
\(\alpha + \beta = -\frac{-4}{2} = 2\)
\(\alpha\beta = \frac{5}{2}\)
(a) \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = (2)^2 - 2\left(\frac{5}{2}\right) = 4 - 5 = -1\)
(b) \(\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{2}{\frac{5}{2}} = \frac{4}{5}\)
(c) \((\alpha - \beta)^2 = (\alpha + \beta)^2 - 4\alpha\beta = (2)^2 - 4\left(\frac{5}{2}\right) = 4 - 10 = -6\)
(d) \(\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha\beta)^2} = \frac{-1}{\left(\frac{5}{2}\right)^2} = \frac{-1}{\frac{25}{4}} = -\frac{4}{25}\)
(e) \(\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 + \beta^2 - \alpha\beta) = (2)(-1 - \frac{5}{2}) = (2)\left(-\frac{7}{2}\right) = -7\)
In simple words: Express each required expression purely in terms of \((\alpha + \beta)\) and \((\alpha\beta)\) using standard algebraic formulas.

Exam Tip: Memorize the cubic sum identity: \(\alpha^3 + \beta^3 = (\alpha + \beta)[(\alpha + \beta)^2 - 3\alpha\beta]\).

Question 10. Obtain all the zeros of the polynomial \(p(x) = 3x^4 - 15x^3 + 17x^2 + 5x - 6\) if two zeroes are \(-\frac{1}{\sqrt{3}}\) and \(\frac{1}{\sqrt{3}}\).
Answer: Given zeros are \(x = -\frac{1}{\sqrt{3}}\) and \(x = \frac{1}{\sqrt{3}}\).
Corresponding factors are \(\left(x + \frac{1}{\sqrt{3}}\right)\) and \(\left(x - \frac{1}{\sqrt{3}}\right)\).
Product of factors = \(\left(x + \frac{1}{\sqrt{3}}\right)\left(x - \frac{1}{\sqrt{3}}\right) = x^2 - \frac{1}{3}\).
For polynomial division, we can use the equivalent factor \(3x^2 - 1\).
Dividing \(3x^4 - 15x^3 + 17x^2 + 5x - 6\) by \(3x^2 - 1\):
Quotient = \(x^2 - 5x + 6\)
Now, factorizing the quotient \(x^2 - 5x + 6\):
\(x^2 - 3x - 2x + 6 = 0\)
\(x(x - 3) - 2(x - 3) = 0\)
\((x - 3)(x - 2) = 0 \implies x = 3, 2\)
Thus, all zeros of the polynomial are \(-\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, 2, 3\).
In simple words: Combine the given rationalized roots into a quadratic divisor, divide the polynomial, and factorize the resulting quotient to get all remaining roots.

Exam Tip: Scaling fractional divisors like \(x^2 - \frac{1}{3}\) to integer coefficients \(3x^2 - 1\) makes polynomial long division much easier.

Question 11. Give examples of polynomials \(p(x)\), \(g(x)\), \(q(x)\) and \(r(x)\) which satisfy the division algorithm. (a) \(\text{deg } p(x) = \text{deg } q(x)\) (b) \(\text{deg } q(x) = \text{deg } r(x)\) (c) \(\text{deg } q(x) = 0\).
Answer: (a) When degree of dividend equals degree of quotient, the divisor must be a non-zero constant:
Let \(p(x) = 2x^2 + 4x + 6\), \(g(x) = 2\), then \(q(x) = x^2 + 2x + 3\), \(r(x) = 0\). Here \(\text{deg } p(x) = 2\) and \(\text{deg } q(x) = 2\).
(b) When degree of quotient equals degree of remainder, degree of divisor is higher:
Let \(p(x) = x^3 + x^2 + x + 1\), \(g(x) = x^2\), then \(q(x) = x + 1\), \(r(x) = x + 1\). Here \(\text{deg } q(x) = 1\) and \(\text{deg } r(x) = 1\).
(c) When degree of quotient is 0, the quotient is a constant:
Let \(p(x) = x^2 + 3x + 5\), \(g(x) = x^2 + 1\), then \(q(x) = 1\), \(r(x) = 3x + 4\). Here \(\text{deg } q(x) = 0\).
In simple words: Construct simple polynomials keeping the degree constraints of the division algorithm \(\text{deg } r(x) < \text{deg } g(x)\) in mind.

Exam Tip: Always verify that the remainder degree is strictly less than the divisor degree in any division example.

Question 12. If the zeros of the polynomial \(ax^3 + 3bx^2 + 3cx + d\) are in AP, Prove that \(2b^3 - 3abc + a^2d = 0\).
Answer: Let the three zeros in Arithmetic Progression be \(\alpha = m - n\), \(\beta = m\), and \(\gamma = m + n\).
Sum of zeros: \(\alpha + \beta + \gamma = (m - n) + m + (m + n) = 3m = -\frac{3b}{a} \implies m = -\frac{b}{a}\).
Since \(m\) is a zero of the polynomial, substituting \(x = -\frac{b}{a}\) into the cubic gives a primary relation, or we use sum of products taken two at a time:
\(\alpha\beta + \beta\gamma + \gamma\alpha = (m - n)m + m(m + n) + (m + n)(m - n) = 3m^2 - n^2 = \frac{3c}{a}\)
Substituting \(m = -\frac{b}{a}\):
\(3\left(-\frac{b}{a}\right)^2 - n^2 = \frac{3c}{a} \implies \frac{3b^2}{a^2} - \frac{3c}{a} = n^2 \implies n^2 = \frac{3b^2 - 3ac}{a^2}\)
Product of zeros: \(\alpha\beta\gamma = (m - n)m(m + n) = m(m^2 - n^2) = -\frac{d}{a}\)
Substituting values of \(m\) and \(n^2\):
\(\left(-\frac{b}{a}\right)\left[\left(-\frac{b}{a}\right)^2 - \frac{3b^2 - 3ac}{a^2}\right] = -\frac{d}{a}\)
Simplifying both sides and multiplying by \(a^3\):
\(2b^3 - 3abc + a^2d = 0\).
In simple words: Assume roots as \(m-n\), \(m\), and \(m+n\), use relations between roots and coefficients, and eliminate \(m\) and \(n\) to get the result.

Exam Tip: Whenever zeros are given in AP, always assume them as \(m-n, m, m+n\) to simplify the sum of roots calculation.

Question 13. Find the number of zeros of the polynomial from the graph given.
Answer: The number of zeros of a polynomial is equal to the total number of points where its graph intersects the x-axis. From the given vertical line graph intersecting the x-axis at exactly one point, the number of zeros is 1.
In simple words: Count how many times the graph crosses the horizontal x-axis.

Exam Tip: Intersection with the y-axis does not count towards the number of polynomial zeros.

Question 14. If one zero of the polynomial \(3x^2 - 8x + 2k + 1\) is seven times the other, find the zeros and the value of \(k\).
Answer: Let the zeros be \(\alpha\) and \(7\alpha\).
Sum of zeros: \(\alpha + 7\alpha = 8\alpha = -\frac{-8}{3} = \frac{8}{3} \implies \alpha = \frac{1}{3}\).
Therefore, the zeros are \(\frac{1}{3}\) and \(7\left(\frac{1}{3}\right) = \frac{7}{3}\).
Product of zeros: \((\alpha)(7\alpha) = 7\alpha^2 = \frac{2k + 1}{3}\)
Substitute \(\alpha = \frac{1}{3}\):
\(7\left(\frac{1}{3}\right)^2 = \frac{2k + 1}{3} \implies 7\left(\frac{1}{9}\right) = \frac{2k + 1}{3}\)
\(\frac{7}{3} = 2k + 1 \implies 2k = \frac{7}{3} - 1 = \frac{4}{3} \implies k = \frac{2}{3}\).
In simple words: Take roots as \(\alpha\) and \(7\alpha\), find \(\alpha\) using the sum of roots formula, and find \(k\) using the product of roots formula.

Exam Tip: Always solve for the roots first using the sum of coefficients before moving to the constant term containing the unknown \(k\).

Question 15. If \((n - k)\) is a factor of the polynomials \(x^2 + px + q\) & \(x^2 + mx + n\). Prove that \(k = n + \frac{n - q}{m - p}\).
Answer: Since \((n - k)\) is a factor of \(x^2 + px + q\), substituting \(x = n - k\) gives zero:
\((n - k)^2 + p(n - k) + q = 0\) --- (1)
Since \((n - k)\) is also a factor of \(x^2 + mx + n\):
\((n - k)^2 + m(n - k) + n = 0\) --- (2)
Equating (1) and (2):
\((n - k)^2 + p(n - k) + q = (n - k)^2 + m(n - k) + n\)
\(p(n - k) - m(n - k) = n - q\)
\((n - k)(p - m) = n - q\)
\(n - k = \frac{n - q}{p - m} = -\frac{n - q}{m - p}\)
\(k = n + \frac{n - q}{m - p}\).
In simple words: Substitute the root \((n-k)\) into both polynomial equations by the Factor Theorem, equate them, and rearrange for \(k\).

Exam Tip: The Factor Theorem states that if \((x - a)\) is a factor of \(f(x)\), then \(f(a) = 0\).

Question 16. If \(2, \frac{1}{2}\) are the zeros of \(px^2 + 5x + r\), prove that \(p = r\).
Answer: Let \(\alpha = 2\) and \(\beta = \frac{1}{2}\) be the zeros of \(px^2 + 5x + r\).
Sum of zeros: \(2 + \frac{1}{2} = \frac{5}{2} = -\frac{5}{p} \implies p = -2\) (Wait, re-evaluating: sum formula is \(-\frac{5}{p} = \frac{5}{2} \implies p = -2\)).
Product of zeros: \((2)\left(\frac{1}{2}\right) = 1 = \frac{r}{p} \implies r = p\).
Thus, \(p = r\).
In simple words: Use the product of roots formula \(\alpha\beta = \frac{r}{p}\) where product equals 1 to directly prove \(p = r\).

Exam Tip: The product of roots \(c/a\) is often the quickest way to establish equality between leading and constant coefficients.

Question 17. If \(m, n\) are zeroes of \(ax^2 - 5x + c\), find the value of \(a\) and \(c\) if \(m + n = mn = 10\).
Answer: For polynomial \(ax^2 - 5x + c\):
Sum of zeros: \(m + n = -\frac{-5}{a} = \frac{5}{a}\)
Product of zeros: \(mn = \frac{c}{a}\)
Given that \(m + n = 10\) and \(mn = 10\):
\(\frac{5}{a} = 10 \implies a = \frac{5}{10} = \frac{1}{2}\)
\(\frac{c}{a} = 10 \implies c = 10a = 10\left(\frac{1}{2}\right) = 5\)
Thus, \(a = \frac{1}{2}\) and \(c = 5\).
In simple words: Equate the standard coefficient formulas for sum and product to the given value 10 to solve for \(a\) and \(c\).

Exam Tip: Always equate \(-\frac{b}{a}\) and \(\frac{c}{a}\) directly to the given numerical values of sum and product.

Question 18. What must be subtracted from \(8x^4 + 14x^3 - 2x^2 + 7x - 8\) so that the resulting polynomial is exactly divisible by \(4x^2 + 3x - 2\).
Answer: Let the polynomial be \(P(x) = 8x^4 + 14x^3 - 2x^2 + 7x - 8\) and divisor be \(g(x) = 4x^2 + 3x - 2\).
Performing long division of \(P(x)\) by \(g(x)\):
1. Divide \(8x^4\) by \(4x^2\) to get \(2x^2\).
2. Multiply and subtract to get remainder terms step-by-step.
The final remainder obtained after complete long division is \(14x - 10\).
To make \(P(x)\) exactly divisible, this remainder must be subtracted.
Therefore, the polynomial to be subtracted is \(14x - 10\).
In simple words: Divide the polynomial by the given quadratic divisor; whatever remainder is left over must be subtracted.

Exam Tip: The polynomial that must be subtracted is always equal to the exact remainder obtained from long division.

Question 19. What must be added to the polynomial \(p(x) = x^4 + 2x^3 - 2x^2 + x - 1\) so that the resulting polynomial is exactly divisible by \(x^2 + 2x - 3\).
Answer: Let the polynomial to be added be \(ax + b\).
First, perform long division of \(p(x)\) by \(x^2 + 2x - 3\):
Dividing \(x^4 + 2x^3 - 2x^2 + x - 1\) by \(x^2 + 2x - 3\) yields a quotient of \(x^2 + 1\) and a remainder of \(-x + 2\).
For the new polynomial to be divisible, the remainder plus the added term must equal zero:
\((-x + 2) + (\text{Polynomial added}) = 0\)
\(\text{Polynomial added} = x - 2\).
In simple words: Find the remainder by long division and take its negative (additive inverse) as the expression that must be added.

Exam Tip: Remember: Subtract the remainder, but add the negative of the remainder (its additive inverse).