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Q2(iii):
Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases:
(iii) $p(x) = x^3 – 4x^2 + x + 6, g(x) = x – 3$
Solution :
Given Variables & Theoretical Foundation
We are given the following polynomial and a linear divisor:
- Dividend Polynomial: $p(x) = x^3 - 4x^2 + x + 6$
- Divisor Polynomial: $g(x) = x - 3$
Theoretical Justification [The Factor Theorem]: The Factor Theorem states that a polynomial $p(x)$ has a factor $(x - c)$ if and only if $p(c) = 0$. Therefore, to determine if $g(x)$ is a factor of $p(x)$, we must evaluate the polynomial at the root of $g(x)$ and check if the remainder is zero.
Step 1: Determining the Root of the Divisor
To find the value of $c$ to substitute into $p(x)$, we set the divisor $g(x)$ equal to zero and solve for $x$:
$g(x) = 0$
$x - 3 = 0$
$x = 3$
Here, our test value is $c = 3$.
Step 2: Evaluating the Polynomial $p(x)$ at $x = 3$
We substitute $x = 3$ into the original polynomial $p(x)$ to find the remainder [Per the Remainder Theorem]:
$p(3) = (3)^3 - 4(3)^2 + (3) + 6$
Now, we perform the arithmetic operations step-by-step:
- Calculate the cube of 3: $(3)^3 = 27$
- Calculate the square of 3, then multiply by -4: $-4(3)^2 = -4(9) = -36$
- Substitute these values back into the expression:
$p(3) = 27 - 36 + 3 + 6$
Group the positive and negative terms to simplify the addition:
$p(3) = (27 + 3 + 6) - 36$
$p(3) = 36 - 36$
$p(3) = 0$
Step 3: Graphical Verification of the Roots
Because $p(3) = 0$, $x = 3$ is an $x$-intercept (or root) of the polynomial curve $y = x^3 - 4x^2 + x + 6$. The graph below plots the exact coordinates of the polynomial, visually confirming that the curve intersects the $x$-axis exactly at $x = 3$.
Step 4: Logical Conclusion
Since the evaluation of the polynomial at $x = 3$ yields a remainder of exactly $0$ ($p(3) = 0$), the condition of the Factor Theorem is perfectly satisfied.
Final Solution: By the Factor Theorem, since $p(3) = 0$, $g(x) = x - 3$ is indeed a factor of the polynomial $p(x) = x^3 - 4x^2 + x + 6$.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.3
- Q1(i): Determine which of the following polynomials has $(x + 1)$ a factor : (i) $x^3 + x^2 + x + 1$
- Q1(ii): Determine which of the following polynomials has $(x + 1)$ a factor : (ii) $x^4 + x^3 + x^2 + x + 1$
- Q1(iii): Determine which of the following polynomials has $(x + 1)$ a factor : (iii) $x^4 + 3x^3 + 3x^2 + x + 1$
- Q1(iv): Determine which of the following polynomials has $(x + 1)$ a factor : (iv) $x^3 – x^2 – (2 + \sqrt{2})x + \sqrt{2}$
- Q2(i): Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (i) $p(x) = 2x^3 + x^2 – 2x – 1, g(x) = x + 1$
- Q2(ii): Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (ii) $p(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 2$
- Q3(i): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (i) $p(x) = x^2 + x + k$
- Q3(ii): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (ii) $p(x) = 2x^2 + kx + \sqrt{2}$
- Q3(iii): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (iii) $p(x) = kx^2 – \sqrt{2}x + 1$
- Q3(iv): Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (iv) $p(x) = kx^2 – 3x + k$
- Q4(i): Factorise : (i) $12x^2 – 7x + 1$
- Q4(ii): Factorise : (ii) $2x^2 + 7x + 3$
- Q4(iii): Factorise : (iii) $6x^2 + 5x – 6$
- Q4(iv): Factorise : (iv) $3x^2 – x – 4$
- Q5(i): Factorise : (i) $x^3 – 2x^2 – x + 2$
- Q5(ii): Factorise : (ii) $x^3 – 3x^2 – 9x – 5$
- Q5(iii): Factorise : (iii) $x^3 + 13x^2 + 32x + 20$
- Q5(iv): Factorise : (iv) $2y^3 + y^2 – 2y – 1$
CBSE Solutions for Class 9 Mathematics Polynomials
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