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Q2(i):
Use the Factor Theorem to determine whether $g(x)$ is a factor of $p(x)$ in each of the following cases:
(i) $p(x) = 2x^3 + x^2 – 2x – 1, g(x) = x + 1$
Solution :
Initial Setup & Theoretical Foundation
We are given two polynomials:
- The dividend polynomial: $p(x) = 2x^3 + x^2 - 2x - 1$
- The divisor polynomial: $g(x) = x + 1$
To determine whether $g(x)$ is a factor of $p(x)$, we utilize the Factor Theorem. [Per the Factor Theorem, a linear polynomial $x - c$ is a factor of a polynomial $p(x)$ if and only if the polynomial evaluated at $c$ equals zero, i.e., $p(c) = 0$.]
Step 1: Determining the Zero of the Divisor Polynomial $g(x)$
First, we must find the root (or zero) of the linear divisor $g(x)$. We do this by setting the polynomial equal to zero and solving for $x$:
$g(x) = 0$
$x + 1 = 0$
$x = -1$
Thus, the value to be substituted into the dividend polynomial $p(x)$ is $c = -1$.
Step 2: Evaluating the Polynomial $p(x)$ at $x = -1$
We now substitute $x = -1$ into the polynomial $p(x)$ to find the remainder. [By the Remainder Theorem, evaluating $p(-1)$ yields the exact remainder of the division of $p(x)$ by $x + 1$.]
$p(-1) = 2(-1)^3 + (-1)^2 - 2(-1) - 1$
Evaluating each term sequentially based on the order of operations (exponentiation first):
- $(-1)^3 = -1 \implies 2(-1)^3 = 2(-1) = -2$
- $(-1)^2 = 1$
- $-2(-1) = 2$
- The constant term remains $-1$
Substituting these evaluated terms back into the equation:
$p(-1) = -2 + 1 + 2 - 1$
Grouping the positive and negative terms to simplify the arithmetic:
$p(-1) = (-2 + 2) + (1 - 1)$
$p(-1) = 0 + 0$
$p(-1) = 0$
Step 3: Applying the Factor Theorem
The evaluation yields $p(-1) = 0$. Because the remainder is exactly zero, the condition of the Factor Theorem is perfectly satisfied. This proves that there is no remainder when $2x^3 + x^2 - 2x - 1$ is divided by $x + 1$.
Final Solution: Since $p(-1) = 0$, by the Factor Theorem, $g(x) = x + 1$ is a factor of the polynomial $p(x) = 2x^3 + x^2 - 2x - 1$.
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(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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