CBSE Class 9 Mathematics Polynomials Assignment Set 10

POLYNOMIALS

Find Remainder (1-5)

Question 1. Divide \(4x^4 - 3x^3 - 2x^2 + x - 7\) by \((x - 1)\)
Answer: Let \(p(x) = 4x^4 - 3x^3 - 2x^2 + x - 7\)
By Remainder Theorem, when \(p(x)\) is divided by \(x - 1\), the remainder is \(p(1)\).
Substitute \(x = 1\) into \(p(x)\):
\(p(1) = 4(1)^4 - 3(1)^3 - 2(1)^2 + (1) - 7\)
\(p(1) = 4 - 3 - 2 + 1 - 7\)
\(p(1) = 5 - 12 = -7\)
Remainder = \(-7\)
In simple words: Substitute \(x = 1\) into the polynomial to get the remainder directly.

Exam Tip: Always state the Remainder Theorem formula before substituting the zero of the divisor.

Question 2. Divide \(3x^4 + 2x^3 - \frac{x^2}{3} + \frac{2}{27}\) by \(\left(x + \frac{2}{3}\right)\)
Answer: Let \(p(x) = 3x^4 + 2x^3 - \frac{x^2}{3} + \frac{2}{27}\)
The zero of the divisor \(x + \frac{2}{3}\) is \(x = -\frac{2}{3}\).
By Remainder Theorem, the remainder is \(p\left(-\frac{2}{3}\right)\):
\(p\left(-\frac{2}{3}\right) = 3\left(-\frac{2}{3}\right)^4 + 2\left(-\frac{2}{3}\right)^3 - \frac{1}{3}\left(-\frac{2}{3}\right)^2 + \frac{2}{27}\)
\(p\left(-\frac{2}{3}\right) = 3\left(\frac{16}{81}\right) + 2\left(-\frac{8}{27}\right) - \frac{1}{3}\left(\frac{4}{9}\right) + \frac{2}{27}\)
\(p\left(-\frac{2}{3}\right) = \frac{48}{81} - \frac{16}{27} - \frac{4}{27} + \frac{2}{27}\)
Simplify fractions with denominator 27: \(\frac{48}{81} = \frac{16}{27}\)
\(p\left(-\frac{2}{3}\right) = \frac{16}{27} - \frac{16}{27} - \frac{4}{27} + \frac{2}{27} = -\frac{2}{27}\)
Remainder = \(-\frac{2}{27}\)
In simple words: Substitute \(x = -\frac{2}{3}\) into the given polynomial and carefully evaluate powers of fractions.

Exam Tip: Be extra careful with signs when substituting negative fractions into even and odd powers.

Question 3. Divide \(x^4 + 1\) by \((x + 1)\)
Answer: Let \(p(x) = x^4 + 1\)
The zero of the divisor \(x + 1\) is \(x = -1\).
By Remainder Theorem, the remainder is \(p(-1)\):
\(p(-1) = (-1)^4 + 1\)
\(p(-1) = 1 + 1 = 2\)
Remainder = 2
In simple words: Substitute \(x = -1\) into \(x^4 + 1\) to find the remainder.

Exam Tip: Remember that any negative number raised to an even power becomes positive.

Question 4. Divide \(3x^4 - 4x^3 - 3x - 1\) by \((x - 1)\)
Answer: Let \(p(x) = 3x^4 - 4x^3 - 3x - 1\)
The zero of the divisor \(x - 1\) is \(x = 1\).
By Remainder Theorem, the remainder is \(p(1)\):
\(p(1) = 3(1)^4 - 4(1)^3 - 3(1) - 1\)
\(p(1) = 3(1) - 4(1) - 3 - 1\)
\(p(1) = 3 - 4 - 3 - 1 = -5\)
Remainder = \(-5\)
In simple words: Substitute \(x = 1\) into the polynomial to compute the remainder.

Question 5. Divide \(x^3 - 1\) by \(x - 1\)
Answer: Let \(p(x) = x^3 - 1\)
The zero of the divisor \(x - 1\) is \(x = 1\).
By Remainder Theorem, the remainder is \(p(1)\):
\(p(1) = (1)^3 - 1\)
\(p(1) = 1 - 1 = 0\)
Remainder = 0
In simple words: Since the remainder is 0, \(x - 1\) is a factor of \(x^3 - 1\).

Exam Tip: A remainder of zero indicates that the divisor completely divides the polynomial.

Question 6. Without actual division, find the remainder when \(9x^3 - 3x^2 + x - 5\) is divided by \(\left(x - \frac{2}{3}\right)\)
Answer: Let \(p(x) = 9x^3 - 3x^2 + x - 5\)
The zero of the divisor \(x - \frac{2}{3}\) is \(x = \frac{2}{3}\).
By Remainder Theorem, remainder = \(p\left(\frac{2}{3}\right)\):
\(p\left(\frac{2}{3}\right) = 9\left(\frac{2}{3}\right)^3 - 3\left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right) - 5\)
\(p\left(\frac{2}{3}\right) = 9\left(\frac{8}{27}\right) - 3\left(\frac{4}{9}\right) + \frac{2}{3} - 5\)
\(p\left(\frac{2}{3}\right) = \frac{72}{27} - \frac{12}{9} + \frac{2}{3} - 5\)
Simplify fractions: \(\frac{72}{27} = \frac{8}{3}\) and \(\frac{12}{9} = \frac{4}{3}\)
\(p\left(\frac{2}{3}\right) = \frac{8}{3} - \frac{4}{3} + \frac{2}{3} - 5\)
Combine terms with common denominator 3:
\(p\left(\frac{2}{3}\right) = \frac{8 - 4 + 2}{3} - 5 = \frac{6}{3} - 5\)
\(p\left(\frac{2}{3}\right) = 2 - 5 = -3\)
Remainder = \(-3\)
In simple words: Use the Remainder Theorem by substituting \(x = \frac{2}{3}\) instead of performing long division.

Exam Tip: Simplify fractions at each intermediate step to avoid large numbers.

Question 7. Using remainder theorem, find the remainder when \(4x^3 - 12x^2 + 14x - 3\) is divided by \(2x - 1\)
Answer: Let \(p(x) = 4x^3 - 12x^2 + 14x - 3\)
To find the zero of the divisor, set \(2x - 1 = 0\), which gives \(x = \frac{1}{2}\).
By Remainder Theorem, remainder = \(p\left(\frac{1}{2}\right)\):
\(p\left(\frac{1}{2}\right) = 4\left(\frac{1}{2}\right)^3 - 12\left(\frac{1}{2}\right)^2 + 14\left(\frac{1}{2}\right) - 3\)
\(p\left(\frac{1}{2}\right) = 4\left(\frac{1}{8}\right) - 12\left(\frac{1}{4}\right) + 7 - 3\)
\(p\left(\frac{1}{2}\right) = \frac{4}{8} - \frac{12}{4} + 4\)
\(p\left(\frac{1}{2}\right) = \frac{1}{2} - 3 + 4\)
\(p\left(\frac{1}{2}\right) = \frac{1}{2} + 1 = \frac{3}{2}\)
Remainder = \(\frac{3}{2}\)
In simple words: Equate the linear divisor to zero to find the fractional value of \(x\) to substitute.

Exam Tip: Always solve \(ax + b = 0\) to find the correct value to substitute for \(x\).

Question 8. Check whether \(7 + 3x\) is a factor of \(3x^2 + 7x\)
Answer: Let \(p(x) = 3x^2 + 7x\)
To find the zero of the divisor \(7 + 3x\), set \(7 + 3x = 0\), which gives \(x = -\frac{7}{3}\).
By Factor Theorem, \(7 + 3x\) is a factor of \(p(x)\) if \(p\left(-\frac{7}{3}\right) = 0\).
Substitute \(x = -\frac{7}{3}\) into \(p(x)\):
\(p\left(-\frac{7}{3}\right) = 3\left(-\frac{7}{3}\right)^2 + 7\left(-\frac{7}{3}\right)\)
\(p\left(-\frac{7}{3}\right) = 3\left(\frac{49}{9}\right) - \frac{49}{3}\)
\(p\left(-\frac{7}{3}\right) = \frac{49}{3} - \frac{49}{3} = 0\)
Since the remainder is 0, \(7 + 3x\) is a factor of \(3x^2 + 7x\).
In simple words: If substituting the root into the polynomial results in zero, then it is a factor.

Exam Tip: Conclude clearly with 'Yes' or 'No' as asked in the question.

Question 9. Find the value of \(k\) for which the polynomial \(2x^4 + 3x^3 + 2kx^2 + 3x + 6\) is completely divisible by \(x + 2\)
Answer: Let \(p(x) = 2x^4 + 3x^3 + 2kx^2 + 3x + 6\)
Since \(p(x)\) is completely divisible by \(x + 2\), by Factor Theorem, the remainder must be 0 when \(x = -2\).
That is, \(p(-2) = 0\).
Substitute \(x = -2\) into \(p(x)\):
\(2(-2)^4 + 3(-2)^3 + 2k(-2)^2 + 3(-2) + 6 = 0\)
\(2(16) + 3(-8) + 2k(4) - 6 + 6 = 0\)
\(32 - 24 + 8k = 0\)
\(8 + 8k = 0\)
\(8k = -8\)
\(k = -1\)
Correction Note: Based on standard assignment solutions, the evaluated result gives \(k = -1\).
In simple words: Set \(p(-2) = 0\) because complete divisibility means the remainder is zero.

Exam Tip: Complete divisibility is synonymous with a remainder of zero.

Question 10. Find the value of \(k\) for which the polynomial \(kx^3 + 9x^2 + 4x - 10\) leaves the remainder as \(-22\) when divided by \(x + 3\)
Answer: Let \(p(x) = kx^3 + 9x^2 + 4x - 10\)
The zero of the divisor \(x + 3\) is \(x = -3\).
Given that the remainder when divided by \(x + 3\) is \(-22\), by Remainder Theorem: \(p(-3) = -22\)
Substitute \(x = -3\) into \(p(x)\):
\(k(-3)^3 + 9(-3)^2 + 4(-3) - 10 = -22\)
\(k(-27) + 9(9) - 12 - 10 = -22\)
\(-27k + 81 - 22 = -22\)
Subtract \(-22\) from both sides:
\(-27k + 81 = 0\)
\(27k = 81\)
\(k = \frac{81}{27} = 3\)
\(k = 3\)
In simple words: Equate the polynomial expression evaluated at \(x = -3\) to the given remainder \(-22\) and solve for \(k\).

Exam Tip: Be careful to equate the substitution result to the given non-zero remainder, not to zero.