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Q3(iv):
Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases:
(iv) $p(x) = kx^2 – 3x + k$
Solution :
Initial Setup & Theoretical Foundation
We are given the polynomial function $p(x)$ and a linear binomial $g(x)$ which is stated to be a factor of $p(x)$:
- Dividend Polynomial: $p(x) = kx^2 - 3x + k$
- Linear Factor: $g(x) = x - 1$
[Per the Factor Theorem, a polynomial $p(x)$ has a factor $(x - c)$ if and only if the polynomial evaluates to zero at $x = c$, meaning $p(c) = 0$. This is a direct corollary of the Remainder Theorem, where a remainder of zero indicates perfect divisibility.]
Step 1: Determining the Root of the Linear Factor
To apply the Factor Theorem, we must first find the zero of the linear divisor $g(x)$. We do this by setting the factor equal to zero and solving for $x$:
$x - 1 = 0$
$x = 1$
[This establishes that $c = 1$. Therefore, if $x - 1$ is a factor of $p(x)$, evaluating $p(x)$ at $x = 1$ must yield exactly $0$.]
Step 2: Applying the Factor Theorem
We substitute $x = 1$ into the polynomial $p(x)$ and set the entire expression equal to zero:
$p(1) = k(1)^2 - 3(1) + k = 0$
Step 3: Algebraic Evaluation for $k$
Now, we simplify the equation to isolate and solve for the unknown constant $k$:
- Evaluate the exponent and multiplication: $k(1) - 3 + k = 0$
- Combine like terms ($k + k$): $2k - 3 = 0$
- Add $3$ to both sides of the equation: $2k = 3$
- Divide by $2$: $k = \frac{3}{2}$
Graphical Verification (Visualizing the Polynomial)
By substituting $k = \frac{3}{2}$ back into the original equation, the polynomial becomes $p(x) = \frac{3}{2}x^2 - 3x + \frac{3}{2}$. Factoring out $\frac{3}{2}$ yields $p(x) = \frac{3}{2}(x^2 - 2x + 1) = \frac{3}{2}(x - 1)^2$. This reveals that $x = 1$ is a repeated root, meaning the parabola's vertex rests exactly on the x-axis at $x = 1$.
Final Solution: The value of $k$ for which $x - 1$ is a factor of $p(x) = kx^2 - 3x + k$ is $k = \frac{3}{2}$.
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(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) $p(x) = kx^2 – \sqrt{2}x + 1$
- 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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