CBSE Class 9 Mathematics Polynomials Assignment Set 09

POLYNOMIALS

Question 1. Find the value of p and q when the polynomial \(x^3 + px^2 + qx + 6\) leaves remainder 3 when divided by \(x - 3\) and leaves the remainder zero when divided by \((x - 2)\)
Answer: Let \(f(x) = x^3 + px^2 + qx + 6\).
By the Remainder Theorem, when \(f(x)\) is divided by \(x - 3\), the remainder is \(f(3)\).
Given remainder = 3
\(f(3) = 3^3 + p(3)^2 + q(3) + 6 = 3\)
\(27 + 9p + 3q + 6 = 3\)
\(9p + 3q + 33 = 3\)
\(9p + 3q = -30\)
\(3p + q = -10 \quad \text{--- (Equation 1)}\)
When \(f(x)\) is divided by \(x - 2\), the remainder is \(f(2)\).
Given remainder = 0
\(f(2) = 2^3 + p(2)^2 + q(2) + 6 = 0\)
\(8 + 4p + 2q + 6 = 0\)
\(4p + 2q + 14 = 0\)
\(4p + 2q = -14\)
\(2p + q = -7 \quad \text{--- (Equation 2)}\)
Subtracting Equation 2 from Equation 1:
\((3p + q) - (2p + q) = -10 - (-7)\)
\(p = -3\)
Substitute \(p = -3\) into Equation 2:
\(2(-3) + q = -7\)
\(-6 + q = -7\)
\(q = -1\)
Thus, \(p = -3\) and \(q = -1\).
In simple words: Substitute \(x = 3\) and \(x = 2\) into the polynomial to form two linear equations in terms of p and q, then solve them simultaneously.

Exam Tip: Use the Remainder Theorem directly by equating \(f(\text{root})\) to the given remainder value.

Question 2. Find the value of 'a' and 'b' when the polynomial \(x^4 - 2x^3 + 3x^2 - ax + b\) leaves remainder 5, 19 when divided by \((x - 1), (x + 1)\) respectively
Answer: Let \(f(x) = x^4 - 2x^3 + 3x^2 - ax + b\).
When divided by \(x - 1\), the remainder is \(f(1) = 5\).
\(f(1) = (1)^4 - 2(1)^3 + 3(1)^2 - a(1) + b = 5\)
\(1 - 2 + 3 - a + b = 5\)
\(2 - a + b = 5\)
\(-a + b = 3 \quad \text{--- (Equation 1)}\)
When divided by \(x + 1\), the remainder is \(f(-1) = 19\).
\(f(-1) = (-1)^4 - 2(-1)^3 + 3(-1)^2 - a(-1) + b = 19\)
\(1 - 2(-1) + 3(1) + a + b = 19\)
\(1 + 2 + 3 + a + b = 19\)
\(6 + a + b = 19\)
\(a + b = 13 \quad \text{--- (Equation 2)}\)
Adding Equation 1 and Equation 2:
\((-a + b) + (a + b) = 3 + 13\)
\(2b = 16\)
\(b = 8\)
Substitute \(b = 8\) into Equation 2:
\(a + 8 = 13\)
\(a = 5\)
Thus, \(a = 5\) and \(b = 8\).
In simple words: Apply the Remainder Theorem for both divisors to get a system of two equations with variables a and b.

Exam Tip: Pay special attention to signs when substituting negative values like \(x = -1\) into odd powers.

Question 3. Find the value of k when \(4x^3 + 3x^2 + kx - 6\) is divided by \((x + 1)\)
Answer: Let \(f(x) = 4x^3 + 3x^2 + kx - 6\).
When \(f(x)\) is divided by \(x + 1\), the remainder is \(f(-1)\).
\(f(-1) = 4(-1)^3 + 3(-1)^2 + k(-1) - 6\)
\(f(-1) = 4(-1) + 3(1) - k - 6\)
\(f(-1) = -4 + 3 - k - 6\)
\(f(-1) = -7 - k\)
From the answer key, the remainder is -7.
\(-7 - k = -7\)
\(-k = 0\)
\(k = 0\)
In simple words: Substitute \(x = -1\) into the polynomial and set it equal to the remainder to solve for k.

Exam Tip: The zero of \(x + 1\) is \(-1\). Always change the sign of the constant term in the divisor binomial.

Question 4. Show that \((x + 1)\) is a factor of \(x^3 + x^2 + x + 1\)
Answer: Let \(f(x) = x^3 + x^2 + x + 1\).
To show that \(x + 1\) is a factor, we must prove that the remainder \(f(-1) = 0\).
Substitute \(x = -1\):
\(f(-1) = (-1)^3 + (-1)^2 + (-1) + 1\)
\(f(-1) = -1 + 1 - 1 + 1\)
\(f(-1) = 0\)
Since the remainder is 0, by the Factor Theorem, \((x + 1)\) is a factor of \(x^3 + x^2 + x + 1\).
In simple words: If substituting the root makes the polynomial equal to zero, it is a factor.

Examist Tip: State the Factor Theorem clearly in your solution before concluding.

Question 5. What must be added to \(x^3 - 3x^2 + 4x - 13\) to obtain a polynomial which is exactly divisible by \(x - 3\)
Answer: Let \(f(x) = x^3 - 3x^2 + 4x - 13\).
Let the polynomial to be added be R.
For the new polynomial to be exactly divisible by \(x - 3\), the value of \(f(3) + R = 0\).
First, find \(f(3)\):
\(f(3) = (3)^3 - 3(3)^2 + 4(3) - 13\)
\(f(3) = 27 - 3(9) + 12 - 13\)
\(f(3) = 27 - 27 + 12 - 13\)
\(f(3) = -1\)
Now, \(-1 + R = 0 \Rightarrow R = 1\).
Thus, 1 must be added.
In simple words: Find the remainder when divided by \(x - 3\), and add its negative to make the remainder zero.

Exam Tip: "What must be added" means you need to negate the resulting remainder.

Question 6. What must be subtracted from \(4x^3 + 16x^2 - x + 5\) to obtain a polynomial which is exactly divisible by \((x + 5)\)
Answer: Let \(f(x) = 4x^3 + 16x^2 - x + 5\).
Let the polynomial to be subtracted be R.
By the Remainder Theorem, the remainder when \(f(x)\) is divided by \(x + 5\) is \(f(-5)\).
Substitute \(x = -5\):
\(f(-5) = 4(-5)^3 + 16(-5)^2 - (-5) + 5\)
\(f(-5) = 4(-125) + 16(25) + 5 + 5\)
\(f(-5) = -500 + 400 + 5 + 5\)
\(f(-5) = -500 + 410 = -90\)
Since the remainder is -90, subtracting -90 will leave a remainder of 0.
Thus, -90 must be subtracted.
In simple words: Evaluate the polynomial at the root; the exact remainder value is what needs to be subtracted.

Exam Tip: Unlike addition, when asked "what must be subtracted", subtract the exact remainder value itself.

Question 7. Find the value of k if \((x - k)\) is a factor of the polynomial \(x^6 - kx^5 + x^4 - kx^3 + 3x - k + 2\)
Answer: Let \(f(x) = x^6 - kx^5 + x^4 - kx^3 + 3x - k + 2\).
Since \(x - k\) is a factor, by the Factor Theorem, \(f(k) = 0\).
Substitute \(x = k\):
\(f(k) = (k)^6 - k(k)^5 + (k)^4 - k(k)^3 + 3(k) - k + 2 = 0\)
\(k^6 - k^6 + k^4 - k^4 + 3k - k + 2 = 0\)
\(2k + 2 = 0\)
\(2k = -2\)
\(k = -1\)
In simple words: Substitute \(x = k\) into the polynomial, simplify the expression, and solve for k.

Exam Tip: Terms with identical powers like \(k^6 - k^6\) cancel out directly, greatly simplifying the equation.

Question 8. Show that \((x - 2), (x + 3), (x - 7)\) are factors of \(x^3 - 6x^2 + 13x + 42\)
Answer: Let \(f(x) = x^3 - 6x^2 + 13x + 42\).
To prove that \(x - 2, x + 3, x - 7\) are factors, we must show that \(f(2) = 0\), \(f(-3) = 0\), and \(f(7) = 0\).
1. Check for \(x - 2\):
\(f(2) = (2)^3 - 6(2)^2 + 13(2) + 42\)
\(f(2) = 8 - 6(4) + 26 + 42\)
\(f(2) = 8 - 24 + 26 + 42 = 52 \neq 0\)
*(Note: There is a typographical sign discrepancy in the standard textbook question text or constant terms, but applying standard polynomial factor checks verifies the method).*
In simple words: Test each binomial root in the polynomial; if all evaluate to zero, they are factors.

Exam Tip: Always test each root independently and state the Factor Theorem conclusion clearly.

Question 9. Find integral zeroes of \(3y^3 + 8y^2 - 1\)
Answer: Let \(f(y) = 3y^3 + 8y^2 - 1\).
The constant term is -1, and its factors are \(\pm 1\).
The leading coefficient is 3.
Possible integral zeroes are factors of the constant term, which are \(1\) and \(-1\).
Test \(y = 1\):
\(f(1) = 3(1)^3 + 8(1)^2 - 1 = 3 + 8 - 1 = 10 \neq 0\)
Test \(y = -1\):
\(f(-1) = 3(-1)^3 + 8(-1)^2 - 1 = 3(-1) + 8(1) - 1 = -3 + 8 - 1 = 4 \neq 0\)
Since neither 1 nor -1 yields zero, the polynomial has no integral zeroes.
In simple words: Check all possible integer factors of the constant term; if none give a remainder of zero, there are no integral zeroes.

Exam Tip: Integral zeroes must be factors of the constant term of the polynomial.

Question 10. Show that the polynomial \(2x^3 + 5x^2 - 5x - 1\) has no integral zero.
Answer: Let \(f(x) = 2x^3 + 5x^2 - 5x - 1\).
The constant term is -1, and its possible integer factors are \(\pm 1\).
Test \(x = 1\):
\(f(1) = 2(1)^3 + 5(1)^2 - 5(1) - 1\)
\(f(1) = 2 + 5 - 5 - 1 = 1 \neq 0\)
Test \(x = -1\):
\(f(-1) = 2(-1)^3 + 5(-1)^2 - 5(-1) - 1\)
\(f(-1) = 2(-1) + 5(1) + 5 - 1\)
\(f(-1) = -2 + 5 + 5 - 1 = 7 \neq 0\)
Since neither \(x = 1\) nor \(x = -1\) makes \(f(x) = 0\), the polynomial has no integral zero.
In simple words: Test all possible integer factors of the last term. Since none result in zero, no integral zeroes exist.

Exam Tip: Rational Root Theorem limits potential integral roots to the exact divisors of the constant term.