Question

If polynomials \[a{x^3} + 3{x^2} - 3\] and \[2{x^3} - 5x + a\] when divided by \[\left( {x - 4} \right)\] leaves remainders as \[{R_1}\] and \[{R_2}\] respectively. Find the values of \[a\] in \[2{R_1} - {R_2} = 0\] .

Answer 1

Hint: Here in this question, we have to find the value of \[a\] in \[2{R_1} - {R_2} = 0\] . The given two polynomials which can divide by polynomial \[\left( {x - 4} \right)\] it leaves two reminders \[{R_1}\] and \[{R_2}\] respectively means by remainder theorem we can find the \[x\] value using divisor, when substitute \[x\] value in the two given polynomials we get \[{R_1}\] and \[{R_2}\] and later substitute \[{R_1}\] and \[{R_2}\] with \[x\] value in \[2{R_1} - {R_2} = 0\] when it equates to zero on by further simplification we get the required solution.

Complete step-by-step answer:
Let us consider the given two polynomials as:
 \[f\left( x \right) = a{x^3} + 3{x^2} - 3\]
 \[g\left( x \right) = 2{x^3} - 5x + a\] and
 \[h\left( x \right) = x - 4\]
Here, we have to find the value of \[a\] in \[2{R_1} - {R_2} = 0\]
Now, we will find the reminders \[{R_1}\] and \[{R_2}\] when the polynomials \[f\left( x \right)\] and \[g\left( x \right)\] respectively are divided by polynomial \[h\left( x \right)\] .
By remainder theorem, when polynomial \[f\left( x \right)\] is divisible by \[h\left( x \right)\] the reminder is \[{R_1} = f\left( 4 \right)\] , then
 \[ \Rightarrow \,\,{R_1} = a{\left( 4 \right)^3} + 3{\left( 4 \right)^2} - 3\]
 \[ \Rightarrow \,\,{R_1} = a\left( {64} \right) + 3\left( {16} \right) - 3\]
 \[ \Rightarrow \,\,{R_1} = 64a + 48 - 3\]
 \[ \Rightarrow \,\,{R_1} = 64a + 45\] ----------(1)
And
Again, by remainder theorem, when polynomial \[g\left( x \right)\] is divisible by \[h\left( x \right)\] the reminder is \[{R_2} = g\left( 4 \right)\] , then
 \[ \Rightarrow \,\,{R_2} = 2{\left( 4 \right)^3} - 5\left( 4 \right) + a\]
 \[ \Rightarrow \,\,{R_2} = 2\left( {64} \right) - 20 + a\]
 \[ \Rightarrow \,\,{R_2} = 128 - 20 + a\]
 \[ \Rightarrow \,\,{R_2} = 108 + a\] ----------(2)
Now to find the value of \[a\]
Consider, the given
 \[ \Rightarrow \,\,2{R_1} - {R_2} = 0\]
On substituting equation (1) and (2), we have
 \[ \Rightarrow \,\,2\left( {64a + 45} \right) - \left( {108 + a} \right) = 0\]
 \[ \Rightarrow \,\,128a + 90 - 108 - a = 0\]
On simplification, we get
 \[ \Rightarrow \,\,127a - 18 = 0\]
Add 18 on both side, we have
 \[ \Rightarrow \,\,127a - 18 + 18 = 0 + 18\]
 \[ \Rightarrow \,\,127a = 18\]
Divide both side by 127, then we get
 \[ \Rightarrow \,\,a = \dfrac{{18}}{{127}}\]
Hence, the value of \[a = \dfrac{{18}}{{127}}\] .
So, the correct answer is “ \[a = \dfrac{{18}}{{127}}\] ”.

Note: The any degree polynomial when divided by linear polynomial or polynomial having degree 1 the remainder can be found easily by using a reminder theorem. The remainder theorem states that: Let \[P\left( x \right)\] be any polynomial of degree greater than or equal to one and let \[a\] be any real number. If \[P\left( x \right)\] is divided by the linear polynomial \[x - a\] , then the remainder is \[P\left( a \right)\] .