
If $x + a$ is factor of ${x^4} - {a^2}{x^2} + 3x - 6a$ then value of $a$=
A) 0
B) $ - 1$
C) 1
D) 2
Answer
543.6k+ views
Hint:
For this type of question, first we have to apply the factor theorem for that consider the polynomial as $f\left( x \right)$, then put the given factor equal to zero, say $x = a$. Then put the value of $x = a$ in a given polynomial such that $f\left( a \right) = 0$. And finally simplify it to get the result.
Complete step by step solution:
To find the value of $a$, let us consider the given polynomial as $f\left( x \right)$, so that
$f\left( x \right)$= ${x^4} - {a^2}{x^2} + 3x - 6a$
We know that if $f\left( x \right)$ is exactly divisible by $x + a$ , then $x + a$ is a factor of $f\left( x \right)$.
Therefore on putting it equal to zero, we get
$x + a = 0$
$ \Rightarrow x = - a$
Now we have to put the above value of $x$ in the given polynomial, we get
$f\left( a \right) = {\left( { - a} \right)^4} - {a^2}\left( { - {a^2}} \right) + 3\left( { - a} \right) - 6a$=$0$
On opening the brackets,
$f\left( a \right) = {a^4} - {a^4} - 3a - 6a = 0$
On cancellation of same and opposite term and addition of same term, we get
$f\left( a \right) = - 9a = 0$
On transferring number one side, we get
$a = 0$
Additional information:
Factor theorem- Let $f\left( x \right)$ be a polynomial of degree $n \geqslant $1 and $a$ be any real number such that
$\left( i \right)$ If $f\left( a \right) = 0,$ then $x - a$ is a factor of $f\left( x \right)$
$\left( {ii} \right)$ If $x - a$ is a factor of $f\left( x \right)$ , then $f\left( a \right) = 0,$
Factor- Multiplying two numbers or terms or any algebraic terms gives a product. The numbers that we multiply are the factors of the product.
Polynomial- An algebraic expression involving a single variable, which has only the whole number as exponent of the variable, are called polynomials in one variable.
Note:
In such types of problems, sometimes students apply Remainder Theorem, which is also correct but it takes too much time and solution will be lengthy and also confusing. So we should take care of these differences of theorems.
For this type of question, first we have to apply the factor theorem for that consider the polynomial as $f\left( x \right)$, then put the given factor equal to zero, say $x = a$. Then put the value of $x = a$ in a given polynomial such that $f\left( a \right) = 0$. And finally simplify it to get the result.
Complete step by step solution:
To find the value of $a$, let us consider the given polynomial as $f\left( x \right)$, so that
$f\left( x \right)$= ${x^4} - {a^2}{x^2} + 3x - 6a$
We know that if $f\left( x \right)$ is exactly divisible by $x + a$ , then $x + a$ is a factor of $f\left( x \right)$.
Therefore on putting it equal to zero, we get
$x + a = 0$
$ \Rightarrow x = - a$
Now we have to put the above value of $x$ in the given polynomial, we get
$f\left( a \right) = {\left( { - a} \right)^4} - {a^2}\left( { - {a^2}} \right) + 3\left( { - a} \right) - 6a$=$0$
On opening the brackets,
$f\left( a \right) = {a^4} - {a^4} - 3a - 6a = 0$
On cancellation of same and opposite term and addition of same term, we get
$f\left( a \right) = - 9a = 0$
On transferring number one side, we get
$a = 0$
Additional information:
Factor theorem- Let $f\left( x \right)$ be a polynomial of degree $n \geqslant $1 and $a$ be any real number such that
$\left( i \right)$ If $f\left( a \right) = 0,$ then $x - a$ is a factor of $f\left( x \right)$
$\left( {ii} \right)$ If $x - a$ is a factor of $f\left( x \right)$ , then $f\left( a \right) = 0,$
Factor- Multiplying two numbers or terms or any algebraic terms gives a product. The numbers that we multiply are the factors of the product.
Polynomial- An algebraic expression involving a single variable, which has only the whole number as exponent of the variable, are called polynomials in one variable.
Note:
In such types of problems, sometimes students apply Remainder Theorem, which is also correct but it takes too much time and solution will be lengthy and also confusing. So we should take care of these differences of theorems.
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