Question

How do you long divide $\dfrac{{6{x^2} + 5x - 4}}{{2x + 3}}$?

Answer 1

Hint: In this question we have to divide the given polynomial with the binomial given, for this we will use long division method, and we should know that the long division method of polynomials where we divide $p\left( x \right)$by$g\left( x \right)$comparing the first terms of the divisor polynomial and the dividend polynomial until we get a remainder $r\left( x \right)$ whose degree is less than the degree $g\left( x \right)$.

Complete step by step solution:
Given polynomial is $6{x^2} + 5x - 4$ and we have to divide this polynomial with the polynomial $2x + 3$, For this, first consider both the leading terms of the dividend and divisor, here they are $6{x^2}$ and $2x$,
Next, divide the leading term of the dividend by the leading term of the divisor.
Next place the partial quotient on right side, i.e., here it is $3x$,
$ \Rightarrow \left. {2x + 3} \right)6{x^2} + 5x - 4\left( {3x} \right.$$ \Rightarrow \left. {2x + 3} \right)6{x^2} + 5x - 4\left( {3x} \right.$,
Now take the partial quotient you placed on the right side i.e.,$3x$, and distribute into the divisor $2x + 3$.
Position the product of $3x$ and $2x + 3$ under the dividend. Make sure to align them by similar terms, i.e.,$6{x^2}$
$ \Rightarrow \left. {2x + 3} \right)6{x^2} + 5x - 4\left( {3x} \right.$
                    $\underline {6{x^2} + 9x} $,
Perform subtraction by switching the signs of the bottom polynomial, i.e, here,
$ \Rightarrow \left. {2x + 3} \right)6{x^2} + 5x - 4\left( {3x} \right.$
                    $\underline {6{x^2} + 9x} $,
                  $\left( - \right)$$\left( - \right)$,
Proceed with regular addition vertically. We will see that the first column from the left cancels each other out, i.e.,
$ \Rightarrow \left. {2x + 3} \right)6{x^2} + 5x - 4\left( {3x} \right.$
                    $\underline {6{x^2} + 9x} $,
                                $ - 4x$
Next, look at the bottom polynomial, $ - 4x$, take its leading term which is $ - 4x$ and divide it by the leading term of the divisor $2x$,
Again, place the partial quotient on right side, i.e.,
$ \Rightarrow \left. {2x + 3} \right)6{x^2} + 5x - 4\left( {3x} \right. - 2$
                    $\underline {6{x^2} + 9x} $,
                                $ - 4x$
Now carry down the next adjacent term of the dividend, i.e., here it will be $ - 4$, now we get,
$ \Rightarrow \left. {2x + 3} \right)6{x^2} + 5x - 4\left( {3x} \right. - 2$
                    $\underline {6{x^2} + 9x} $,
                           $ - 4x - 4$
Use the partial quotient that you put up, $ - 2$ and distribute into the divisor, i.e., Place the product of $ - 2$ and leading term of the divisor i.e.,$2x$, we get,
$ \Rightarrow \left. {2x + 3} \right)6{x^2} + 5x - 4\left( {3x} \right. - 2$
                    $\underline {6{x^2} + 9x} $,
                           $ - 4x - 4$
$6{x^2} + 5x - 4 = \left( {2x + 3} \right)\left( {3x - 2} \right) + 2$ ..
Now subtract the terms we get,
$ \Rightarrow \left. {2x + 3} \right)6{x^2} + 5x - 4\left( {3x} \right. - 2$
                    $\underline {6{x^2} + 9x} $,
                    $ - 4x - 4$
      $\underline { - 4x - 6} $
                                      $2$
So, here the remainder is $2$, and the quotient is $3x - 2$.

$\therefore $ The quotient when the polynomial $6{x^2} + 5x - 4$ is divided by $2x + 3$ will be equal to$3x - 2$ and the remainder is $2$, and the given polynomial can be written as $6{x^2} + 5x - 4 = \left( {2x + 3} \right)\left( {3x - 2} \right) + 2$.

Note:
When we divide a dividend polynomial$p\left( x \right)$with degree$n$by some divisor$g\left( x \right)$with degree $m$, then$m \leqslant n$then we get the quotient polynomial$q\left( x \right)$of degree $n - m$ and the remainder polynomial as $r\left( x \right)$of degree $h$, then$h < m$.