Question

If \[{x^2} + ax + 1\]is a factor of \[a{x^3} + bx + c\], then
A. \[b + a - {a^2} = 0,a = c\]
B. \[b - a + {a^2} = 0,a = c\]
C. \[b + a - {a^2} = 0,a = 0\]
D. None of these

Answer 1

Hint: Here we use the given factor to divide the equation and then write it in \[a = bq + r\] form where \[a\] is the dividend, \[b\] is the divisor, \[q\] is the quotient and \[r\] is remainder.
* If there is one factor of an equation, then there can be other factors of the same equation as well which when multiplied to each other give us the original equation.
* Factor of an equation means that equation is divisible by that given factor.
* Method of long division is used to find factors of a polynomial where the polynomial is arranged in descending order of degree of the variable and the variable having no coefficient is either written as zero or an empty space is left there at its place.

Complete Step-by-step answer:
Given, \[{x^2} + ax + 1\] is a factor of \[a{x^3} + bx + c\].
We have to find the remaining factors of \[a{x^3} + bx + c\] when divide by \[{x^2} + ax + 1\] .
Perform division taking \[{x^2} + ax + 1\] as divisor and \[a{x^3} + bx + c\] as dividend.

                              \[ax - {a^2}\]
\[{x^2} + ax + 1)\overline {a{x^3} + bx + c} \]
                          \[
  \underline { - a{x^3} - ax - {a^2}{x^2}} \\
                       0 - {a^2}{x^2} + (b - a)x + c \\
 \]
                                \[
    {a^2}{x^2} + {a^3}x + {a^2} \\
    \overline {0 + (b - a + {a^3})x + (c + {a^2})} \\
 \]

\[a{x^3} + bx + c\] can be written as \[a{x^3} + bx + c = (ax - {a^2})({x^2} + ax + 1) + \left( {b - a + {a^3}} \right)x + \left( {c + {a^2}} \right)\]

For \[{x^2} + ax + 1\] to be a factor of \[a{x^3} + bx + c\], \[\left( {b - a + {a^2}} \right)x + \left( {c + {a^2}} \right)\] must be equal to 0.
\[\left( {b - a + {a^2}} \right)x + \left( {c + {a^2}} \right)\] will be equal to 0 only when both the terms are individually equal to zero because one of them is a coefficient of \[x\] and other is constant term, that is \[\left( {b - a + {a^2}} \right)x = 0\] and \[\left( {c + {a^2}} \right) = 0\]

Simplify to obtain the result.
\[\left( {b - a + {a^2}} \right) = 0\] and \[c = - {a^2}\]

Option D is correct.

Note: This type of questions can be solved by performing the long division but care should be taken when changing the sign when subtracting the divisor. Students are likely to make a mistake while doing long division so it is better to check your answer by substituting the values in the equation \[a = bq + r\].