2 Answers

DanJ · Answer 1 · 2021-06-07T22:40:36+0000

Final answer:

The quotient when 4x^3 + 2x + 7 is divided by x + 3 is 4x^2 - 10x + 32 with a remainder of -89.

Step-by-step explanation:

To find the quotient when 4x^3 + 2x + 7 is divided by x + 3, we can use long division. Divide the first term, 4x^3, by x which gives 4x^2. Multiply x + 3 by 4x^2 to get 4x^3 + 12x^2. Subtract this from the original expression to get -10x^2 + 2x + 7. Then, divide -10x^2 by x which gives -10x. Multiply x + 3 by -10x to get -10x^2 - 30x. Subtract this from the previous result to get 32x + 7. Finally, divide 32x by x which gives 32. Multiply x + 3 by 32 to get 32x + 96. Subtract this from the previous result to get -89. Therefore, the quotient when 4x^3 + 2x + 7 is divided by x + 3 is 4x^2 - 10x + 32 with a remainder of -89.

Ronapelbaum · Answer 2 · 2021-06-10T22:20:11+0000

Answer:

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

Step-by-step explanation:

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

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

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
,
, and
.

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

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 = -12 .

Again, rewrite the right-hand side:

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