Answer
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Hint-In this question, we use the concept of division algorithm for polynomials. Division algorithm for polynomials states that, suppose f(x) and g(x) are the two polynomials, where $g\left( x \right) \ne 0$ , we can write: \[f\left( x \right) = q\left( x \right) \times g\left( x \right) + r\left( x \right)\] which is same as the \[{\text{Dividend }} = {\text{ Divisor }} \times {\text{ Quotient + Remainder}}\] and where r(x) is the remainder polynomial and degree of r(x) < degree g(x).
Complete step-by-step answer:
Given that Dividend, \[f\left( x \right) = 3{x^5} - 8{x^4} - 5{x^3} + 26{x^2} - 33x + 26\]
Divisor, $g\left( x \right) = {x^3} - 2{x^2} - 4x + 8$
We know that \[f\left( x \right) = q\left( x \right) \times g\left( x \right) + r\left( x \right)\] , where q(x) and r(x) are the quotient and remainder respectively when polynomial f(x) is divided by polynomial g(x).
Now, we know degree of q(x) = degree of f(x)- degree of g(x)
Degree of q(x) =5-3=2
So, the degree of q(x) is 2.
Now we assume, $q\left( x \right) = A{x^2} + Bx + C$
\[
\Rightarrow q\left( x \right) \times g\left( x \right) = \left( {A{x^2} + Bx + C} \right) \times \left( {{x^3} - 2{x^2} - 4x + 8} \right) \\
\Rightarrow A{x^5} + \left( {B - 2A} \right){x^4} + \left( { - 4A - 2B + C} \right){x^3} + \left( {8A - 4B - 2C} \right){x^2} + \left( {8B - 4C} \right)x + 8C \\
\]
We know the degree of r(x) < degree g(x).
Now, we compare the coefficient of degree 5, 4 and 3 with f(x).
After compare we get A=3, B-2A=-8 and -4A-2B+C=-5
A=3, B=-2 and C=3.
So, the quotient $q\left( x \right) = 3{x^2} - 2x + 3$
Now, we use \[f\left( x \right) = q\left( x \right) \times g\left( x \right) + r\left( x \right)\]
$
\Rightarrow r\left( x \right) = f\left( x \right) - q\left( x \right) \times g\left( x \right) \\
\Rightarrow r\left( x \right) = 3{x^5} - 8{x^4} - 5{x^3} + 26{x^2} - 33x + 26 - q\left( x \right) \times g\left( x \right) \\
$
Put the value of $q\left( x \right) \times g\left( x \right)$ after substituting the value of A, B and C.
$
\Rightarrow r\left( x \right) = 3{x^5} - 8{x^4} - 5{x^3} + 26{x^2} - 33x + 26 - 3{x^5} + 8{x^4} + 5{x^3} - 26{x^2} - 28x - 24 \\
\Rightarrow r\left( x \right) = - 5x + 2 \\
$
So, the remainder is $r\left( x \right) = - 5x + 2$ .
Note-In such types of questions we generally face problems to find the quotient so we use some simple tips to find quotients in an easy way. First we find the degree of quotient and remainder by using the formula mentioned in above and then find the product of quotient and divisor. Then compare the coefficient of the higher degree like 5, 4 and 3 with dividend. So, we will get the quotient.
Complete step-by-step answer:
Given that Dividend, \[f\left( x \right) = 3{x^5} - 8{x^4} - 5{x^3} + 26{x^2} - 33x + 26\]
Divisor, $g\left( x \right) = {x^3} - 2{x^2} - 4x + 8$
We know that \[f\left( x \right) = q\left( x \right) \times g\left( x \right) + r\left( x \right)\] , where q(x) and r(x) are the quotient and remainder respectively when polynomial f(x) is divided by polynomial g(x).
Now, we know degree of q(x) = degree of f(x)- degree of g(x)
Degree of q(x) =5-3=2
So, the degree of q(x) is 2.
Now we assume, $q\left( x \right) = A{x^2} + Bx + C$
\[
\Rightarrow q\left( x \right) \times g\left( x \right) = \left( {A{x^2} + Bx + C} \right) \times \left( {{x^3} - 2{x^2} - 4x + 8} \right) \\
\Rightarrow A{x^5} + \left( {B - 2A} \right){x^4} + \left( { - 4A - 2B + C} \right){x^3} + \left( {8A - 4B - 2C} \right){x^2} + \left( {8B - 4C} \right)x + 8C \\
\]
We know the degree of r(x) < degree g(x).
Now, we compare the coefficient of degree 5, 4 and 3 with f(x).
After compare we get A=3, B-2A=-8 and -4A-2B+C=-5
A=3, B=-2 and C=3.
So, the quotient $q\left( x \right) = 3{x^2} - 2x + 3$
Now, we use \[f\left( x \right) = q\left( x \right) \times g\left( x \right) + r\left( x \right)\]
$
\Rightarrow r\left( x \right) = f\left( x \right) - q\left( x \right) \times g\left( x \right) \\
\Rightarrow r\left( x \right) = 3{x^5} - 8{x^4} - 5{x^3} + 26{x^2} - 33x + 26 - q\left( x \right) \times g\left( x \right) \\
$
Put the value of $q\left( x \right) \times g\left( x \right)$ after substituting the value of A, B and C.
$
\Rightarrow r\left( x \right) = 3{x^5} - 8{x^4} - 5{x^3} + 26{x^2} - 33x + 26 - 3{x^5} + 8{x^4} + 5{x^3} - 26{x^2} - 28x - 24 \\
\Rightarrow r\left( x \right) = - 5x + 2 \\
$
So, the remainder is $r\left( x \right) = - 5x + 2$ .
Note-In such types of questions we generally face problems to find the quotient so we use some simple tips to find quotients in an easy way. First we find the degree of quotient and remainder by using the formula mentioned in above and then find the product of quotient and divisor. Then compare the coefficient of the higher degree like 5, 4 and 3 with dividend. So, we will get the quotient.
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