
The polynomial \[a{x^3} + b{x^2} + x - 6\] has \[\left( {x + 2} \right)\] as a factor and leaves a remainder 4 when divided by \[\left( {x - 2} \right)\]. Find \[a\] and \[b\].
A.\[a = 0\]
B.\[b = 2\]
C.\[a = 2\]
D.\[b = 0\]
Answer
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Hint: Here, we will use the factor theorem \[p\left( { - 2} \right) = 0\], as \[\left( {x + 2} \right)\] is a factor of polynomial \[f\left( x \right)\] and then when \[p\left( x \right)\] is divided by \[x - 2\], the remainder is 4, using the remainder theorem\[p\left( 2 \right) = 4\]. Then we will simplify the equations to find the required value.
Complete step by step Answer:
We are given that the polynomial \[a{x^3} + b{x^2} + x - 6\] has \[\left( {x + 2} \right)\] as a factor and leaves a remainder 4 when divided by \[\left( {x - 2} \right)\].
Let us assume that \[p\left( x \right) = a{x^3} + b{x^2} + x - 6\].
Since \[\left( {x + 2} \right)\] is a factor of polynomial \[p\left( x \right)\], then by factor theorem \[p\left( { - 2} \right) = 0\].
Replacing \[ - 2\] for \[x\] in the given polynomial and taking it equal to 0, we get
\[
\Rightarrow a{\left( { - 2} \right)^3} + b{\left( { - 2} \right)^2} + \left( { - 2} \right) - 6 = 0 \\
\Rightarrow - 8a + 4b - 8 = 0 \\
\Rightarrow - 2a + b = 2{\text{ ......eq.(1)}} \\
\]
Also when \[p\left( x \right)\] is divided by \[x - 2\], the remainder is 4, using the remainder theorem\[p\left( 2 \right) = 4\], we get
\[
\Rightarrow a{\left( 2 \right)^3} + b{\left( 2 \right)^2} + 2 - 6 = 4 \\
\Rightarrow 8a + 4b + 2 - 6 = 4 \\
\Rightarrow 8a + 4b = 8 \\
\Rightarrow 2a + b = 2{\text{ ......eq.(2)}} \\
\]
Adding equation (1) and equation (2), we get
\[
\Rightarrow \left( { - 2a + b} \right) + \left( {2a + b} \right) = 2 + 2 \\
\Rightarrow 2b = 4 \\
\Rightarrow b = 2 \\
\]
Substituting the value of \[b\] in the equation (1), we get
\[
\Rightarrow - 2a + 2 = 2 \\
\Rightarrow - 2a = 0 \\
\Rightarrow a = 0 \\
\]
Thus, the value of \[a\] is 0 and \[b\] is 2.
Hence, options A and B are correct.
Note: In solving these types of questions, students should be careful while calculations. We can also find the remainder by using the division algorithm method, by dividing the first term of the quotient with the highest term of the dividend, which is really time-consuming and is used when asked in the question. Since in the question we are asked to find using the remainder, so students are advised to solve this question using the remainder theorem.
Complete step by step Answer:
We are given that the polynomial \[a{x^3} + b{x^2} + x - 6\] has \[\left( {x + 2} \right)\] as a factor and leaves a remainder 4 when divided by \[\left( {x - 2} \right)\].
Let us assume that \[p\left( x \right) = a{x^3} + b{x^2} + x - 6\].
Since \[\left( {x + 2} \right)\] is a factor of polynomial \[p\left( x \right)\], then by factor theorem \[p\left( { - 2} \right) = 0\].
Replacing \[ - 2\] for \[x\] in the given polynomial and taking it equal to 0, we get
\[
\Rightarrow a{\left( { - 2} \right)^3} + b{\left( { - 2} \right)^2} + \left( { - 2} \right) - 6 = 0 \\
\Rightarrow - 8a + 4b - 8 = 0 \\
\Rightarrow - 2a + b = 2{\text{ ......eq.(1)}} \\
\]
Also when \[p\left( x \right)\] is divided by \[x - 2\], the remainder is 4, using the remainder theorem\[p\left( 2 \right) = 4\], we get
\[
\Rightarrow a{\left( 2 \right)^3} + b{\left( 2 \right)^2} + 2 - 6 = 4 \\
\Rightarrow 8a + 4b + 2 - 6 = 4 \\
\Rightarrow 8a + 4b = 8 \\
\Rightarrow 2a + b = 2{\text{ ......eq.(2)}} \\
\]
Adding equation (1) and equation (2), we get
\[
\Rightarrow \left( { - 2a + b} \right) + \left( {2a + b} \right) = 2 + 2 \\
\Rightarrow 2b = 4 \\
\Rightarrow b = 2 \\
\]
Substituting the value of \[b\] in the equation (1), we get
\[
\Rightarrow - 2a + 2 = 2 \\
\Rightarrow - 2a = 0 \\
\Rightarrow a = 0 \\
\]
Thus, the value of \[a\] is 0 and \[b\] is 2.
Hence, options A and B are correct.
Note: In solving these types of questions, students should be careful while calculations. We can also find the remainder by using the division algorithm method, by dividing the first term of the quotient with the highest term of the dividend, which is really time-consuming and is used when asked in the question. Since in the question we are asked to find using the remainder, so students are advised to solve this question using the remainder theorem.
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