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Q3(ii):
Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases:
(ii) $p(x) = 2x^2 + kx + \sqrt{2}$
Solution :
Given Variables & Initial Setup
We are given a quadratic polynomial $p(x)$ with an unknown coefficient $k$, and a linear binomial $g(x)$ which is stated to be a factor of $p(x)$.
- The Polynomial: $p(x) = 2x^2 + kx + \sqrt{2}$
- The Linear Divisor: $g(x) = x - 1$
Step 1: Determining the Zero of the Divisor
To apply polynomial remainder and factor theorems, we must first find the root (or zero) of the linear divisor $g(x)$. [Per the Fundamental Theorem of Algebra, a linear polynomial of the form $ax + b$ has exactly one real root].
We set the divisor equal to zero:
$x - 1 = 0$
$x = 1$
The zero of the divisor is $x = 1$.
Step 2: Application of the Factor Theorem
The Factor Theorem states that a polynomial $p(x)$ has a factor $(x - c)$ if and only if $p(c) = 0$. [This is a direct corollary of the Polynomial Remainder Theorem, which states that the remainder of $p(x)$ divided by $(x - c)$ is $p(c)$].
| Divisor Condition | Mathematical Translation | Logical Conclusion |
|---|---|---|
| $(x - a)$ is a factor of $p(x)$ | $p(a) = 0$ | The polynomial divides perfectly with a remainder of $0$. |
| $(x - a)$ is NOT a factor of $p(x)$ | $p(a) \neq 0$ | The division yields a non-zero remainder equal to $p(a)$. |
Since the problem explicitly states that $(x - 1)$ is a factor of $p(x)$, we can definitively establish the following equation:
$p(1) = 0$
Step 3: Algebraic Substitution and Evaluation
We now substitute $x = 1$ into the original polynomial expression $p(x) = 2x^2 + kx + \sqrt{2}$ and equate the entire expression to zero.
$p(1) = 2(1)^2 + k(1) + \sqrt{2} = 0$
Evaluate the exponential and multiplicative terms [By the identity property of multiplication, $1^2 = 1$ and $k \cdot 1 = k$]:
$2(1) + k + \sqrt{2} = 0$
$2 + k + \sqrt{2} = 0$
Step 4: Isolating the Unknown Variable $k$
To find the value of $k$, we must isolate it on one side of the equation. We achieve this by subtracting $2$ and $\sqrt{2}$ from both sides of the equation [Per the Subtraction Property of Equality].
$k = -2 - \sqrt{2}$
For mathematical elegance and standard algebraic formatting, we can factor out the negative sign from the binomial on the right side:
$k = -(2 + \sqrt{2})$
Final Solution: The value of $k$ is $-(2 + \sqrt{2})$.
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(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(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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