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Q1(ii):
Determine which of the following polynomials has $(x + 1)$ a factor :
(ii) $x^4 + x^3 + x^2 + x + 1$
Solution :
Initial Setup & Theoretical Foundation
We are given the polynomial expression:
$P(x) = x^4 + x^3 + x^2 + x + 1$
We must determine whether the linear binomial $D(x) = x + 1$ is a factor of $P(x)$.
[Per the Factor Theorem, a polynomial $P(x)$ has a factor $(x - c)$ if and only if the polynomial evaluated at $c$ equals zero, i.e., $P(c) = 0$. This is a direct corollary of the Remainder Theorem, which states that the remainder of $P(x)$ divided by $(x - c)$ is $P(c)$.]
Step 1: Determining the Zero of the Divisor
To apply the Factor Theorem, we first find the root (or zero) of the linear divisor $D(x)$. We do this by setting the divisor equal to zero and solving for $x$:
$x + 1 = 0$
$x = -1$
Thus, our test value is $c = -1$.
Step 2: Applying the Factor Theorem (Substitution)
We substitute $x = -1$ into the original polynomial $P(x)$ to find the remainder:
$P(-1) = (-1)^4 + (-1)^3 + (-1)^2 + (-1) + 1$
[Theoretical Justification: When a negative number is raised to an even power, the result is positive. When raised to an odd power, the result is negative. Specifically, $(-1)^{2k} = 1$ and $(-1)^{2k+1} = -1$ for any integer $k$.]
- Fourth degree term: $(-1)^4 = 1$
- Cubic term: $(-1)^3 = -1$
- Quadratic term: $(-1)^2 = 1$
- Linear term: $(-1)^1 = -1$
- Constant term: $1$
Step 3: Evaluating the Polynomial
Now, we sum the evaluated terms sequentially:
$P(-1) = 1 + (-1) + 1 + (-1) + 1$
$P(-1) = (1 - 1) + (1 - 1) + 1$
$P(-1) = 0 + 0 + 1$
$P(-1) = 1$
Step 4: Logical Conclusion
The remainder of the division is $1$. Because $P(-1) \neq 0$, the condition required by the Factor Theorem is not satisfied. Therefore, the polynomial does not divide evenly by $(x + 1)$.
Final Solution: Since $P(-1) = 1 \neq 0$, $(x + 1)$ is not a factor of the polynomial $x^4 + x^3 + x^2 + x + 1$.
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(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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