Polynomial Exercise 2.3 Latest NCERT Book Solutions with Practice Questions from RD Sharma
Class 9 Mathematics: Polynomials
Exercise 2.3 Solutions
1. Determine if \((x + 1)\) is a factor:
(i) \(x^3 + x^2 + x + 1\)
Step 1: Use Factor Theorem: If \((x + 1)\) is a factor, then \(p(-1) = 0\).
Step 2: Substitute \(x = -1\):
\[
(-1)^3 + (-1)^2 + (-1) + 1
= -1 + 1 – 1 + 1 = 0
\]
Answer: Yes, \((x + 1)\) is a factor.
(ii) \(x^4 + x^3 + x^2 + x + 1\)
Step 1: Substitute \(x = -1\):
\[
(-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1
= 1 – 1 + 1 – 1 + 1 = 1
\]
Answer: No, \((x + 1)\) is not a factor.
(iii) \(x^4 + 3x^3 + 3x^2 + x + 1\)
Step 1: Substitute \(x = -1\):
\[
(-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
\]
Answer: No, \((x + 1)\) is not a factor.
(iv) \(x^3 – x^2 – (2\sqrt{2})x + \sqrt{2}\)
Step 1: Substitute \(x = -1\):
\[
(-1)^3 – (-1)^2 – (2\sqrt{2})(-1) + \sqrt{2}
= -1 – 1 + 2\sqrt{2} + \sqrt{2} = -2 + 3\sqrt{2}
\]
Answer: No, \((x + 1)\) is not a factor.
2. Use Factor Theorem to check if \(g(x)\) is a factor of \(p(x)\):
(i) \(p(x) = 2x^3 + x^2 – 2x – 1\), \(g(x) = x + 1\)
Step 1: Substitute \(x = -1\) in \(p(x)\):
\[
2(-1)^3 + (-1)^2 – 2(-1) – 1
= -2 + 1 + 2 – 1 = 0
\]
Answer: Yes, \(g(x)\) is a factor.
(ii) \(p(x) = x^3 + 3x^2 + 3x + 1\), \(g(x) = x + 2\)
Step 1: Substitute \(x = -2\) in \(p(x)\):
\[
(-2)^3 + 3(-2)^2 + 3(-2) + 1
= -8 + 12 – 6 + 1 = -1
\]
Answer: No, \(g(x)\) is not a factor.
(iii) \(p(x) = x^3 – 4x^2 + x + 6\), \(g(x) = x – 3\)
Step 1: Substitute \(x = 3\) in \(p(x)\):
\[
3^3 – 4(3)^2 + 3 + 6 = 27 – 36 + 3 + 6 = 0
\]
Answer: Yes, \(g(x)\) is a factor.
3. Find the value of \(k\) if \((x – 1)\) is a factor:
(i) \(p(x) = x^2 + x + k\)
Step 1: Substitute \(x = 1\) in \(p(x)\):
\[
1^2 + 1 + k = 0 \implies 2 + k = 0
\]
Answer: \(k = -2\).
(ii) \(p(x) = 2x^2 + kx + \sqrt{2}\)
Step 1: Substitute \(x = 1\) in \(p(x)\):
\[
2(1)^2 + k(1) + \sqrt{2} = 0
\implies 2 + k + \sqrt{2} = 0
\]
Answer: \(k = -2 – \sqrt{2}\).
(iii) \(p(x) = kx^2 – \sqrt{2}x + 1\)
Step 1: Substitute \(x = 1\) in \(p(x)\):
\[
k(1)^2 – \sqrt{2}(1) + 1 = 0
\implies k – \sqrt{2} + 1 = 0
\]
Answer: \(k = \sqrt{2} – 1\).
(iv) \(p(x) = kx^2 – 3x + k\)
Step 1: Substitute \(x = 1\) in \(p(x)\):
\[
k(1)^2 – 3(1) + k = 0
\implies 2k – 3 = 0
\]
Answer: \(k = \frac{3}{2}\).
4. Factorise:
(i) \(12x^2 – 7x + 1\)
Step 1: Split the middle term:
\(12x^2 – 4x – 3x + 1 = 4x(3x – 1) – 1(3x – 1)\)
Answer: \((4x – 1)(3x – 1)\).
(ii) \(2x^2 + 7x + 3\)
Step 1: Split the middle term:
\(2x^2 + 6x + x + 3
= 2x(x + 3) + 1(x + 3)\)
Answer: \((2x + 1)(x + 3)\).
(iii) \(6x^2 + 5x – 6\)
Step 1: Split the middle term:
\(6x^2 + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)\)
Answer: \((3x – 2)(2x + 3)\).
(iv) \(3x^2 – x – 4\)
Step 1: Split the middle term:
\(3x^2 – 4x + 3x – 4 = x(3x – 4) + 1(3x – 4)\)
Answer: \((x + 1)(3x – 4)\).
5. Factorise:
(i) \(x^3 – 2x^2 – x + 2\)
Step 1: Factor by grouping:
\(x^2(x – 2) – 1(x – 2) = (x^2 – 1)(x – 2)\)
Answer: \((x – 1)(x + 1)(x – 2)\).
(ii) \(x^3 – 3x^2 – 9x – 5\)
Step 1: Use Factor Theorem to find roots.
Answer: \((x + 1)(x – 5)(x + 1)\).
(iii) \(x^3 + 13x^2 + 32x + 20\)
Step 1: Factor by grouping.
Answer: \((x + 1)(x + 2)(x + 10)\).
(iv) \(2y^3 + y^2 – 2y – 1\)
Step 1: Factor by grouping.
Answer: \((y – 1)(2y + 1)(y + 1)\).
Additional Practice Questions (RD Sharma)
1. Verify if \((x – 2)\) is a factor of \(p(x) = x^3 – 4x^2 + x + 6\).
Answer: Yes, \(p(2) = 0\).
2. Find the value of \(k\) if \((x – 1)\) is a factor of \(p(x) = x^3 + kx^2 – 4x + 1\).
Answer: Substitute \(x = 1\) in \(p(x)\):
\(1 + k – 4 + 1 = 0 \implies k = 2\).
3. Factorise \(x^3 – 6x^2 + 11x – 6\).
Answer: \((x – 1)(x – 2)(x – 3)\).
4. Verify if \((x + 3)\) is a factor of \(p(x) = x^3 + 6x^2 + 11x + 6\).
Answer: Yes, \(p(-3) = 0\).
5. Find the zero of \(p(x) = 3x – 7\).
Answer: Solve \(3x – 7 = 0\) ⇒ \(x = \frac{7}{3}\).
6. Factorise \(x^3 – 3x^2 – 9x – 5\).
Answer: \((x + 1)(x – 5)(x + 1)\).
7. Verify if \((x – 1)\) is a factor of \(p(x) = x^4 – 2x^3 + 3x^2 – 4x + 2\).
Answer: No, \(p(1) = 0\).
8. Find the value of \(k\) if \((x – 2)\) is a factor of \(p(x) = x^3 – 3x^2 + kx – 8\).
Answer: Substitute \(x = 2\) in \(p(x)\):
\(8 – 12 + 2k – 8 = 0 \implies k = 6\).
9. Factorise \(x^3 + 13x^2 + 32x + 20\).
Answer: \((x + 1)(x + 2)(x + 10)\).
10. Verify if \((x + 2)\) is a factor of \(p(x) = x^3 + 4x^2 + x – 6\).
Answer: Yes, \(p(-2) = 0\).
