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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:

$p(1) = 0$

Input Root x = 1 Polynomial p(x) 2x² + kx + √2 Remainder 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


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