Question

Divide \[6{x^4} + 13{x^3} + 39{x^2} + 37x + 45\] by \[3{x^2} + 2x + 9\]

Answer 1

Hint: Here, we will use the method of long division and divide the given Dividend by the Divisor to get the required quotient and the remainder. The division is a method that is inverse of the multiplication method. In division, the number that is to be divided is known as dividend and the number by which it is divided is called the divisor.

Complete step-by-step answer:
We will use the method of long division to find the required quotient and remainder.
According to the question, Dividend is \[6{x^4} + 13{x^3} + 39{x^2} + 37x + 45\] and the Divisor is \[3{x^2} + 2x + 9\].
Hence, by using long division method:

                                  \[2x^2+3x+5\]
\[3x^2+2x+9)\overline {6x^4+13x^3+39x^2+37x+45\,\,\,\,\,\,} \]
                 \[
\,\,\,\,\,\, \underline {-\left(6x^4+4x^3+18x^2\right)\,\,\,\,\,\,\,\,} \\
\,\,\,\,\,\,\,\,\,\, \,0+9x^3+21x^2+37x+45 \\
 \]
                         \[
   \underline{-\left(9x^3+6x^2+27x \right)} \\
    \,0+15x^2+10x+45 \\
 \]
                                \[
   \underline{-\left(15x^2+10x+45 \right)} \\
\,\,\,\,\,\,\,\, \,0 \\
 \]


Hence, the quotient is \[2{x^2} + 3x + 5\] and the remainder is 0.
Therefore, we can write this Dividend in the form of:
Dividend \[ = \] Divisor \[ \times \] Quotient\[ + \] Remainder
Substituting the values in the above equation, we get
\[ \Rightarrow 6{x^4} + 13{x^3} + 39{x^2} + 37x + 45 = \left( {3{x^2} + 2x + 9} \right)\left( {2{x^2} + 3x + 5} \right) + 0\]
Now, when we will multiply the terms in RHS, we will get the LHS i.e. the Dividend. This shows that we have divided the dividend by the divisor correctly.
Therefore, when we divide \[6{x^4} + 13{x^3} + 39{x^2} + 37x + 45\] by \[3{x^2} + 2x + 9\], we get the quotient as \[2{x^2} + 3x + 5\] and the remainder as 0.
Hence, this is the required answer.

Note: We will check whether we have found the correct quotient or not by using the formula discussed earlier.
Dividend \[ = \]Divisor \[ \times \] Quotient \[ + \] Remainder
Substituting the values in this equation,
\[ \Rightarrow 6{x^4} + 13{x^3} + 39{x^2} + 37x + 45 = \left( {3{x^2} + 2x + 9} \right)\left( {2{x^2} + 3x + 5} \right) + 0\]
Now, solving for RHS by opening the brackets,
RHS \[ = 3{x^2}\left( {2{x^2} + 3x + 5} \right) + 2x\left( {2{x^2} + 3x + 5} \right) + 9\left( {2{x^2} + 3x + 5} \right)\]
RHS \[ = 6{x^4} + 9{x^3} + 15{x^2} + 4{x^3} + 6{x^2} + 10x + 18{x^2} + 27x + 45\]
Hence, solving further, we get
RHS \[ = 6{x^4} + 13{x^3} + 39{x^2} + 37x + 45\]
Clearly, RHS \[ = \] LHS
Hence, we have verified that our answer is correct.
Therefore, we can always check for our answer using this formula.