If $x - 2$ is a factor of each of the following two polynomials, find the values of $a$ in each case:$x^3 - 2ax^2 + ax - 1$
Given:
Given expression is $x^3 - 2ax^2 + ax - 1$
$x - 2$ is a factor of $x^3 - 2ax^2 + ax - 1$.
To do:
We have to find the value of $a$.
Solution:
We know that,
If $(x-m)$ is a root of $f(x)$ then $f(m)=0$.
Therefore,
$f(2)=0$
$\Rightarrow (2)^3 - 2a(2)^2 + a(2) - 1=0$
$\Rightarrow 8-8a+2a-1=0$
$\Rightarrow 7-6a=0$
$\Rightarrow 6a=7$
$\Rightarrow a=\frac{7}{6}$
The value of $a$ is $\frac{7}{6}$.
Related Articles
- If $x - 2$ is a factor of each of the following two polynomials, find the values of $a$ in each case:$x^5 - 3x^4 - ax^3 + 3ax^2 + 2ax + 4$
- In each of the following two polynomials, find the values of $a$, if $x - a$ is a factor:$x^6 - ax^5 + x^4-ax^3 + 3x-a + 2$
- In each of the following, two polynomials, find the value of $a$, if $x + a$ is a factor.$x^3 + ax^2 - 2x + a + 4$
- In each of the following two polynomials, find the values of $a$, if $x - a$ is a factor:$x^5 - a^2x^3 + 2x + a + 1$
- Using factor theorem, factorize each of the following polynomials:$x^3 + 2x^2 - x - 2$
- Using factor theorem, factorize each of the following polynomials:$x^3 - 2x^2 - x + 2$
- In each of the following, two polynomials, find the value of $a$, if $x + a$ is a factor.$x^4 - a^2x^2 + 3x - a$
- Find value of $a$ if $x+1$ is a factor of $x^3-ax^2+6x-a$.
- Determine if the following polynomials has $(x+1)$ a factor: $x^{3}+x^{2}+x+1$
- Using factor theorem, factorize each of the following polynomials:$3x^3 - x^2 - 3x + 1$
- Find the values of $x$ in each of the following:\( 5^{2 x+3}=1 \)
- Factorize each of the following polynomials:$x^3 + 13x^2 + 31x - 45$ given that $x + 9$ is a factor.
- If $x+2$ is a factor of $x^2+ax+8$, then find the value of $a$.
- Find the values of $x$ in each of the following:\( 5^{x-2} \times 3^{2 x-3}=135 \)
- Determine which of the following polynomials has \( (x+1) \) a factor:(i) \( x^{3}+x^{2}+x+1 \)(ii) \( x^{4}+x^{3}+x^{2}+x+1 \)
Kickstart Your Career
Get certified by completing the course
Get Started
To Continue Learning Please Login
Login with Google