Question

Find the values of a and b if \[{x^2} - 4\] is a factor of \[a{x^4} + 2{x^3} - 3{x^2} + bx - 4\].

Answer 1

Hint:Here we use the given value of the factor of the polynomial and perform a long division process which will give us a remainder. For a polynomial to be a factor of another polynomial, it has to divide the polynomial completely to give the remainder zero.Using this concept we try to find the values of a and b .

Complete step-by-step answer:
A polynomial of degree \[n\] is a polynomial where the variable has highest power \[n\]. It can be written as \[a{x^n} + b{x^{n - 1}} + ..... + cx + d = 0\]
If \[x - a\] is one of the factors of a polynomial \[a{x^n} + b{x^{n - 1}} + ..... + cx + d = 0\] then we divide the polynomial by \[x - a\] to find other factors.
Long division method: when dividing \[a{x^n} + b{x^{n - 1}} + ....c\] by \[px + q\] we perform as
\[px + q)\overline {a{x^n} + b{x^{n - 1}} + ....c} ((a/p){x^{n - 1}} + ...\]
            \[\underline { - a{x^n} + (qa/p){x^{n - 1}}} \]
                          \[0.{x^n} + (b - qa/p){x^{n - 1}}\]
Here we multiply the divisor with such term that gives us the exact same term as the highest power in the dividend and then we proceed in the same way.
We multiply the divisor with such a factor so we cancel out the highest power of the variable in it.

We are given the factor of the polynomial \[a{x^4} + 2{x^3} - 3{x^2} + bx - 4\] is \[{x^2} - 4\].
So, the factor of the polynomial must divide the polynomial completely to give the remainder zero.
Using long division method
\[{x^2} - 4)\overline {a{x^4} + 2{x^3} - 3{x^2} + bx - 4} (a{x^2} + 2x + (4a - 3)\]
            \[\underline {a{x^4} - 4a{x^2}} \]
                       \[2{x^3} + (4a - 3){x^2} + bx - 4\]
                      \[\underline {2{x^3} - 8x} \]
                                  \[(4a - 3){x^2} + (b + 8)x - 4\]
                                  \[\underline {(4a - 3){x^2} - 16a + 12} \]
                                                          \[(b + 8)x - 4 + 16a - 12\]
Since we know when a is divided by b, then the quotient is q and remainder is r. We write the equation as \[a = bq + r\]
Therefore, from long division method we write
\[a{x^4} + 2{x^3} - 3{x^2} + bx - 4 = ({x^2} - 4) \times (a{x^2} + 2x + (4a - 3)) + ((b + 8)x - 4 + 16a - 12)\]
If \[{x^2} - 4\] is a factor of \[a{x^4} + 2{x^3} - 3{x^2} + bx - 4\] then the remainder has to be zero.
Here the remainder is \[(b + 8)x - 4 + 16a - 12\]
We equate the remainder to zero.
\[ \Rightarrow (b + 8)x - 4 + 16a - 12 = 0\]
Solving LHS
\[ \Rightarrow (b + 8)x + 16a - 16 = 0\]
We know that RHS can be written as \[0x + 0\] in the form of a polynomial.
\[ \Rightarrow (b + 8)x + 16a - 16 = 0x + 0\]
We know that the coefficient of x will be equal on both sides and constants will be equal on both sides.
Equating coefficient of x
\[ \Rightarrow b + 8 = 0\]
Shift constant to one side of the equation.
\[ \Rightarrow b = - 8\]
Equating constants
\[ \Rightarrow 16a - 16 = 0\]
Shift constant to one side of the equation.
\[ \Rightarrow 16a = 16\]
Cancel same value from both sides
\[ \Rightarrow a = 1\]
So, the value of a is 1 and b is -8.

Note:Students are likely to make mistakes in the process of long division as they forget to change the sign before the next step. Also keep in mind that the degree of the divisor will always be less than the degree of the dividend, then only it can be a factor of the dividend.