How do you divide \[\dfrac{{4{x^3} - 12{x^2} + 5x + 3}}{{x + 3}}\] ?

Hint: We will take \[x + 3\] as the common term for the polynomial and then solve accordingly to find the quotient and remainder respectively. One can try the long division approach but that may be cumbersome to a first try.

Complete step-by-step solution:
We start with our given polynomial \[4{x^3} - 12{x^2} + 5x + 3\] and take \[x + 3\] common from each of the terms and then settle the terms accordingly.
Take \[(x + 3)\] common from the first term i.e. \[(x + 3)4{x^2}\] so we need to subtract the extra term \[12{x^2}\]
Then, \[(x + 3)(4{x^2} - 12{x^2} - 12{x^2} + 5x + 3)\]
That equals to \[(x + 3)(4{x^2} - 24{x^2} + 5x + 3)\]
Taking \[(x + 3)\]common from the second term,\[ - (x + 3)24x\], so we need to add the extra term \[72x\]
That is, \[(x + 3)4{x^2} - 24x(x + 3) + 72x + 5x + 3\]\[ = (x + 3)4{x^2} - 24x(x + 3) + 77x + 3\]
Again take \[(x + 3)\] common from the third term, \[(x + 3)77\] so we will subtract the extra \[228\]
Therefore finally, \[(x + 3)4{x^2} - 24x(x + 3) + 77(x + 3) - 228\]
\[ = (x + 3)(4{x^2} - 24x + 77) - 228\]
Hence, \[\dfrac{{(x + 3)(4{x^2} - 24x + 77) - 228}}{{x + 3}}\]
\[ \Rightarrow (4{x^2} - 24x + 77) - \dfrac{{228}}{{x + 3}}\]
Then, the remainder is \[ - 228\] and the quotient is\[(4{x^2} - 24x + 77)\].
Additional information: We can also use a polynomial long division method for this kind of problem.
First let us arrange the terms in the decreasing order of their indices (if required). Pen down the missing terms with the coefficients as zero.
Then, for the first term of the quotient, divide the first term of the dividend by the first term of the divisor and then multiply this term of the quotient by the divisor to get the product.
Thereafter, subtract this product from the dividend, and bring down the next term (if any). The difference and the term that is taken down to the bottom from above will form the new dividend. Continue this process until you get a remainder, which can be zero or of a lower index than that of the divisor.

Note: Long division method is sometimes clumsy and cumbersome for computation and so the method of taking the common term and then adjusting the polynomial gives an easier approach to us. The common term that is the denominator gets cancelled out after taking common and the remainder remains if not completely divisible.