Question

What is the quotient when 4x3 + 2x + 7 is divided by x + 3?

Answer 1

Answer:

The quotient of this division is $(4x^2 -12x + 38)$. The remainder here would be $-26$.

Step-by-step explanation:

The numerator $4x^3 + 2x + 7$ is a polynomial about $x$ with degree $3$.

The divisor $x + 3$ is a polynomial, also about $x$, but with degree $1$.

By the division algorithm, the quotient should be of degree $3 - 1 = 2$, while the remainder shall be of degree $1 - 1 = 0$ (i.e., the remainder would be a constant.) Let the quotient be $a\,x^2 + b\, x + c$ with coefficients $a$, $b$, and $c$.

$4x^3 + 2x + 7 = \left(a\,x^2 + b\, x + c\right)(x + 3)$.

Start by finding the first coefficient of the quotient.

The degree-three term on the left-hand side is $4 x^3$. On the right-hand side, that would be $a\, x^3$. Hence $a = 4$.

Now, given that $a = 4$, rewrite the right-hand side:

$\begin{aligned}&\left(4\,x^2 + b\, x + c\right)(x + 3) \cr =& \left(4x^2 + (b\, x + c)\right)(x + 3) \cr =& 4x^2(x + 3) + (bx + c)(x + 3) \cr =& 4x^3 + 12x^2 + (bx + c)(x + 3)\end{aligned}$.

Hence:

$4x^3 + 2x + 7 = 4x^3 + 12x^2 + (b\,x + c)(x + 3)$

Subtract $\left(4x^3 + 12x^2\right$ from both sides of the equation:

$-12x^2 + 2x + 7 = (b\,x + c)(x + 3)$.

The term with a degree of two on the left-hand side has coefficient $(-12)$. Since the only term on the right hand side with degree two would have coefficient $b$, $b = -12$.

Again, rewrite the right-hand side:

$\begin{aligned}&\left(-12 x + c\right)(x + 3) \cr =& \left(-12 x+ c\right)(x + 3) \cr =& (-12x)(x + 3) + c(x + 3) \cr =& -12x^2 -36x + (bx + c)(x + 3)\end{aligned}$.

Subtract $-12x^2 -36x$ from both sides of the equation:

$38x + 7 = c(x + 3)$.

By the same logic, $c = 38$.

Hence the quotient would be $(4x^2 - 12x + 38)$.