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Q1(i):
Determine which of the following polynomials has $(x + 1)$ a factor :
(i) $x^3 + x^2 + x + 1$
Solution :
Initial Setup & Theoretical Foundation
We are tasked with determining whether the linear polynomial $D(x) = x + 1$ is a factor of the given cubic polynomial $P(x) = x^3 + x^2 + x + 1$.
[Per the Factor Theorem of Polynomials], for any polynomial $P(x)$ of degree $n \ge 1$ and any real number $a$, the linear binomial $(x - a)$ is a factor of $P(x)$ if and only if $P(a) = 0$. This theorem is a direct corollary of the Remainder Theorem, which states that dividing $P(x)$ by $(x - a)$ yields a remainder equal to $P(a)$. If the remainder is zero, the divisor is a perfect factor.
Step 1: Determining the Zero of the Divisor
To apply the Factor Theorem, we must first find the root (or zero) of the divisor $D(x)$. We do this by setting the divisor equal to zero and solving for $x$:
$x + 1 = 0$
$x = -1$
Here, our value for $a$ is $-1$. We must now evaluate $P(-1)$.
Step 2: Evaluating the Polynomial at $x = -1$
We substitute $x = -1$ into the original polynomial $P(x) = x^3 + x^2 + x + 1$.
$P(-1) = (-1)^3 + (-1)^2 + (-1) + 1$
To ensure absolute precision, we break down the exponentiation of the negative integer:
| Term | Substitution | Algebraic Expansion | Evaluated Result |
|---|---|---|---|
| $x^3$ | $(-1)^3$ | $(-1) \times (-1) \times (-1)$ | $-1$ |
| $x^2$ | $(-1)^2$ | $(-1) \times (-1)$ | $1$ |
| $x$ | $(-1)^1$ | $-1$ | $-1$ |
| Constant | $1$ | $1$ | $1$ |
Substituting these evaluated terms back into the polynomial equation:
$P(-1) = (-1) + 1 + (-1) + 1$
$P(-1) = 0$
Step 3: Graphical Verification
Graphically, the real roots of a polynomial correspond to the $x$-intercepts of its graph. Because $P(-1) = 0$, the graph of $y = x^3 + x^2 + x + 1$ must intersect the $x$-axis exactly at the coordinate $(-1, 0)$.
Step 4: Analytical Conclusion
Because the evaluation of the polynomial at $x = -1$ yields exactly $0$, the remainder of the division $\frac{x^3 + x^2 + x + 1}{x + 1}$ is $0$. Therefore, the polynomial divides perfectly without any fractional remainder.
Final Solution: Since $P(-1) = 0$, by the Factor Theorem, $(x + 1)$ is indeed a factor of the polynomial $x^3 + x^2 + x + 1$.
More Questions from Class 9 Mathematics Polynomials EXERCISE 2.3
- 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) $p(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 2$
- 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
Chapters in CBSE - Class 9 Mathematics
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