Answer:
The quotient of this division is [tex](4x^2 -12x + 38)[/tex]. The remainder here would be [tex]-26[/tex].
Step-by-step explanation:
The numerator [tex]4x^3 + 2x + 7[/tex] is a polynomial about [tex]x[/tex] with degree [tex]3[/tex].
The divisor [tex]x + 3[/tex] is a polynomial, also about [tex]x[/tex], but with degree [tex]1[/tex].
By the division algorithm, the quotient should be of degree [tex]3 - 1 = 2[/tex], while the remainder shall be of degree [tex]1 - 1 = 0[/tex] (i.e., the remainder would be a constant.) Let the quotient be [tex]a\,x^2 + b\, x + c[/tex] with coefficients [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex].
[tex]4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)[/tex].
Start by finding the first coefficient of the quotient.
The degree-three term on the left-hand side is [tex]4 x^3[/tex]. On the right-hand side, that would be [tex]a\, x^3[/tex]. Hence [tex]a = 4[/tex].
Now, given that [tex]a = 4[/tex], rewrite the right-hand side:
[tex]\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}[/tex].
Hence:
[tex]4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)[/tex]
Subtract [tex]\left(4x^3 + 12x^2\right[/tex] from both sides of the equation:
[tex]-12x^2 + 2x + 7 = (b\,x + c)(x + 3)[/tex].
The term with a degree of two on the left-hand side has coefficient [tex](-12)[/tex]. Since the only term on the right hand side with degree two would have coefficient [tex]b[/tex], [tex]b = -12[/tex].
Again, rewrite the right-hand side:
[tex]\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}[/tex].
Subtract [tex]-12x^2 -36x[/tex] from both sides of the equation:
[tex] 38x + 7 = c(x + 3)[/tex].
By the same logic, [tex]c = 38[/tex].
Hence the quotient would be [tex](4x^2 - 12x + 38)[/tex].
