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Q1(iv):
Determine which of the following polynomials has $(x + 1)$ a factor :
(iv) $x^3 – x^2 – (2 + \sqrt{2})x + \sqrt{2}$
Solution :
Initial Setup & Theoretical Foundation
We are tasked with determining whether the linear polynomial $(x + 1)$ is a factor of the given cubic polynomial:
$P(x) = x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$
[Per the Factor Theorem], a polynomial $D(x) = x - c$ is a factor of a polynomial $P(x)$ if and only if the polynomial evaluated at $c$ yields a remainder of zero, i.e., $P(c) = 0$. This is a direct corollary of the Remainder Theorem, which states that dividing $P(x)$ by $(x - c)$ leaves a remainder equal to $P(c)$.
Step 1: Determining the Root of the Divisor
To apply the Factor Theorem, we must first find the zero (or root) of the divisor $D(x) = x + 1$. We do this by setting the divisor equal to zero:
$x + 1 = 0$
$x = -1$
Thus, our test value is $c = -1$. We must evaluate $P(-1)$.
Step 2: Applying the Factor Theorem
Substitute $x = -1$ into the original polynomial $P(x)$:
$P(-1) = (-1)^3 - (-1)^2 - (2 + \sqrt{2})(-1) + \sqrt{2}$
Step 3: Algebraic Evaluation & Simplification
We will now expand and simplify each term of the expression systematically:
- First term: $(-1)^3 = -1$ [Since an odd power of a negative number remains negative]
- Second term: $-(-1)^2 = -(1) = -1$ [Since an even power of a negative number is positive, but the term is preceded by a negative sign]
- Third term: $-(2 + \sqrt{2})(-1) = +(2 + \sqrt{2}) = 2 + \sqrt{2}$ [The product of two negative signs yields a positive sign]
- Fourth term: $+\sqrt{2}$ [Remains unchanged]
Substituting these simplified terms back into the equation for $P(-1)$:
$P(-1) = -1 - 1 + 2 + \sqrt{2} + \sqrt{2}$
Combine the rational (integer) terms and the irrational terms separately:
$P(-1) = (-1 - 1 + 2) + (\sqrt{2} + \sqrt{2})$
$P(-1) = (-2 + 2) + 2\sqrt{2}$
$P(-1) = 0 + 2\sqrt{2}$
$P(-1) = 2\sqrt{2}$
Step 4: Final Analysis & Conclusion
According to the Factor Theorem, $(x + 1)$ is a factor of $P(x)$ if and only if $P(-1) = 0$. Our algebraic evaluation demonstrates that:
$P(-1) = 2\sqrt{2} \neq 0$
Because the remainder is non-zero, the polynomial $P(x)$ is not perfectly divisible by $(x + 1)$.
Final Solution: Since $P(-1) = 2\sqrt{2} \neq 0$, the linear polynomial $(x + 1)$ is NOT a factor of $x^3 - x^2 - (2 + \sqrt{2})x + \sqrt{2}$.
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) $x^4 + 3x^3 + 3x^2 + x + 1$
- 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) $p(x) = kx^2 – \sqrt{2}x + 1$
- 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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