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Q1(iii):
Determine which of the following polynomials has $(x + 1)$ a factor :
(iii) $x^4 + 3x^3 + 3x^2 + x + 1$
Solution :
Given Variables & Initial Setup
We are tasked with determining whether the linear binomial $D(x) = x + 1$ is a factor of the given quartic polynomial:
$P(x) = x^4 + 3x^3 + 3x^2 + x + 1$
To solve this, we utilize the Factor Theorem. [Per the Factor Theorem of Polynomials, a linear polynomial $(x - c)$ is a factor of a polynomial $P(x)$ if and only if $P(c) = 0$].
Step 1: Determining the Root of the Divisor
First, we must find the zero (or root) of the divisor $D(x) = x + 1$. We do this by equating the divisor to zero and solving for $x$:
$x + 1 = 0$
$x = -1$
[By the properties of equality, subtracting 1 from both sides isolates $x$, yielding the test value $c = -1$].
Step 2: Applying the Factor Theorem
We now substitute the root $x = -1$ into the original polynomial $P(x)$ to evaluate the remainder. [According to the Remainder Theorem, evaluating $P(-1)$ yields the exact remainder of the division $\frac{P(x)}{x+1}$].
$P(-1) = (-1)^4 + 3(-1)^3 + 3(-1)^2 + (-1) + 1$
Step 3: Algebraic Evaluation
We systematically evaluate each term of the polynomial, adhering strictly to the order of operations (exponents before multiplication):
- First term: $(-1)^4 = 1$ [An even power of a negative number results in a positive value].
- Second term: $3(-1)^3 = 3(-1) = -3$ [An odd power of a negative number remains negative].
- Third term: $3(-1)^2 = 3(1) = 3$ [An even power yields a positive, multiplied by 3].
- Fourth term: $(-1) = -1$
- Fifth term: $1 = 1$ [Constant term remains unchanged].
Substituting these evaluated terms back into the polynomial expression:
$P(-1) = 1 - 3 + 3 - 1 + 1$
Step 4: Final Arithmetic Simplification
We now sum the terms sequentially from left to right:
$P(-1) = (1 - 3) + 3 - 1 + 1$
$P(-1) = -2 + 3 - 1 + 1$
$P(-1) = 1 - 1 + 1$
$P(-1) = 0 + 1$
$P(-1) = 1$
Step 5: Final Analysis & Conclusion
The calculated remainder is $1$. Because the remainder is non-zero ($P(-1) \neq 0$), the polynomial $P(x)$ is not perfectly divisible by $(x + 1)$. [By the logical contrapositive of the Factor Theorem, if $P(c) \neq 0$, then $(x - c)$ cannot be a factor].
Final Solution: Since $P(-1) = 1 \neq 0$, the binomial $(x + 1)$ is NOT a factor of the polynomial $x^4 + 3x^3 + 3x^2 + x + 1$.
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(ii): Determine which of the following polynomials has $(x + 1)$ a factor : (ii) $x^4 + x^3 + x^2 + x + 1$
- 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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