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Q3(i):
Find the value of $k$, if $x – 1$ is a factor of $p(x)$ in each of the following cases:
(i) $p(x) = x^2 + x + k$
Solution :
Initial Setup & Theoretical Foundation
We are given the quadratic polynomial:
$p(x) = x^2 + x + k$
We are also given the condition that the linear binomial $(x - 1)$ is a factor of $p(x)$. Our objective is to determine the precise value of the unknown constant $k$ that satisfies this condition.
Step 1: Applying the Factor Theorem
To solve for $k$, we utilize the Factor Theorem. [Per the Factor Theorem of Polynomials, a linear polynomial $(x - a)$ is a factor of a polynomial $p(x)$ if and only if the polynomial evaluated at $x = a$ is equal to zero, i.e., $p(a) = 0$].
To find the root of the divisor, we set the linear factor to zero:
$x - 1 = 0 \implies x = 1$
Therefore, for $(x - 1)$ to be a factor of $p(x)$, the condition $p(1) = 0$ must hold true.
Step 2: Evaluating the Polynomial at the Root
We substitute $x = 1$ into the given polynomial $p(x)$:
$p(1) = (1)^2 + (1) + k$
Simplifying the numerical terms [By applying standard exponentiation and addition]:
$p(1) = 1 + 1 + k$
$p(1) = 2 + k$
Step 3: Solving for the Unknown Constant $k$
According to the condition established in Step 1, we equate $p(1)$ to $0$:
$2 + k = 0$
Isolating $k$ by subtracting $2$ from both sides of the equation [Per the Subtraction Property of Equality]:
$k = -2$
Graphical Verification of the Polynomial
By substituting $k = -2$ back into the original expression, we obtain the complete polynomial: $p(x) = x^2 + x - 2$. Geometrically, the roots of this polynomial correspond to the x-intercepts of its parabolic graph. As shown below, the parabola crosses the x-axis exactly at $x = 1$, visually confirming that $(x - 1)$ is a factor.
Final Solution: The value of $k$ for which $(x - 1)$ is a factor of $p(x) = x^2 + x + k$ is $k = -2$.
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(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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