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Q2(ii):
Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases:
(ii) $p(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 2$
Solution :
Given Variables & Initial Setup
We are tasked with determining whether the linear polynomial $g(x)$ is a factor of the cubic polynomial $p(x)$. The given functions are:
- Dividend Polynomial: $p(x) = x^3 + 3x^2 + 3x + 1$
- Divisor Polynomial: $g(x) = x + 2$
Theoretical Foundation [The Factor Theorem]: The Factor Theorem states that for any polynomial $p(x)$ of degree $n \ge 1$ and any real number $c$, the linear binomial $(x - c)$ is a factor of $p(x)$ if and only if $p(c) = 0$. In other words, the remainder upon dividing $p(x)$ by $(x - c)$ must be exactly zero.
Step 1: Determining the Root of the Divisor $g(x)$
To apply the Factor Theorem, we must first find the zero (or root) of the divisor $g(x)$. We do this by setting $g(x)$ equal to zero and solving for $x$:
$g(x) = 0$
$x + 2 = 0$
$x = -2$
Here, our test value is $c = -2$. According to the Factor Theorem, $g(x)$ will be a factor of $p(x)$ if and only if $p(-2) = 0$.
Step 2: Evaluating the Polynomial $p(x)$ at $x = -2$
We substitute $x = -2$ into the polynomial $p(x)$ to calculate the remainder.
$p(-2) = (-2)^3 + 3(-2)^2 + 3(-2) + 1$
Now, we resolve each term sequentially following the order of operations (exponents, then multiplication):
- Cubic term: $(-2)^3 = -8$
- Quadratic term: $3(-2)^2 = 3(4) = 12$
- Linear term: $3(-2) = -6$
- Constant term: $1$
Substituting these evaluated terms back into the equation yields:
$p(-2) = -8 + 12 - 6 + 1$
Combining the terms from left to right:
$p(-2) = 4 - 6 + 1$
$p(-2) = -2 + 1$
$p(-2) = -1$
Step 3: Analytical Insight & Verification (Algebraic Identity)
To demonstrate rigorous mathematical fluency, we can verify this result by recognizing the structure of $p(x)$. The polynomial $x^3 + 3x^2 + 3x + 1$ is the standard binomial expansion of $(x + 1)^3$ [Per the algebraic identity $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$].
Rewriting $p(x)$:
$p(x) = (x + 1)^3$
Evaluating at $x = -2$:
$p(-2) = (-2 + 1)^3 = (-1)^3 = -1$
This confirms our previous arithmetic calculation with absolute certainty.
Step 4: Graphical Representation of the Remainder
The graph below illustrates the function $p(x) = (x+1)^3$. The Factor Theorem dictates that for $x+2$ to be a factor, the graph must cross the x-axis at $x = -2$. As shown, at $x = -2$, the function evaluates to $-1$, visually confirming a non-zero remainder.
Step 5: Conclusion
Because the evaluation of the polynomial at $x = -2$ yields a remainder of $-1$ rather than $0$ (i.e., $p(-2) \neq 0$), the condition required by the Factor Theorem is not satisfied.
Final Solution: Since $p(-2) = -1 \neq 0$, by the Factor Theorem, $g(x) = x + 2$ is NOT a factor of $p(x) = x^3 + 3x^2 + 3x + 1$.
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(iii): Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases: (iii) $p(x) = x^3 – 4x^2 + x + 6, g(x) = x – 3$
- 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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