Q3(iii):

Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases: (iii) $p(x) = kx^2 – \sqrt{2}x + 1$

Initial Setup & Theoretical Foundation

We are given the quadratic polynomial:

$p(x) = kx^2 - \sqrt{2}x + 1$

We are also given a linear binomial, which acts as a divisor:

$g(x) = x - 1$

The objective is to determine the exact value of the unknown coefficient $k$ under the strict condition that $g(x)$ is a factor of $p(x)$.

Step 1: Application of the Factor Theorem

[Per the Factor Theorem of Polynomials], a linear polynomial of the form $x - c$ is a factor of a polynomial $p(x)$ if and only if the polynomial evaluates to zero at $x = c$. Mathematically, this is expressed as:

$p(c) = 0$

By comparing our given divisor $g(x) = x - 1$ to the standard form $x - c$, we identify the root of the divisor as:

$x - 1 = 0 \implies x = 1$

Therefore, for $x - 1$ to be a factor of $p(x)$, the polynomial must satisfy the condition:

$p(1) = 0$

Step 2: Substitution and Algebraic Manipulation

We substitute $x = 1$ into the original polynomial $p(x)$ to establish our equation:

• $p(1) = k(1)^2 - \sqrt{2}(1) + 1$
• $p(1) = k(1) - \sqrt{2} + 1$
• $p(1) = k - \sqrt{2} + 1$

Equating this expression to zero [as mandated by the Factor Theorem]:

$k - \sqrt{2} + 1 = 0$

Step 3: Solving for the Unknown Constant $k$

To isolate $k$, we transpose the constant terms to the right side of the equation. When moving terms across the equals sign, their signs invert [by the properties of equality]:

• Add $\sqrt{2}$ to both sides: $k + 1 = \sqrt{2}$
• Subtract $1$ from both sides: $k = \sqrt{2} - 1$

Graphical Verification & Geometric Interpretation

Substituting $k = \sqrt{2} - 1$ back into the original equation yields the specific polynomial:

$p(x) = (\sqrt{2} - 1)x^2 - \sqrt{2}x + 1$

Geometrically, the fact that $x - 1$ is a factor means that the parabola represented by $y = p(x)$ must intersect the x-axis exactly at the coordinate $(1, 0)$. The graph below plots this precise quadratic function, visually confirming the root at $x = 1$.

Final Solution: The value of $k$ that makes $x - 1$ a factor of the polynomial $p(x) = kx^2 - \sqrt{2}x + 1$ is $k = \sqrt{2} - 1$.