# Divide 4x^3 + 2x^2 + 10x - 7 by (x-5).

• Oct 18th 2010, 03:51 AM
ejaykasai
Divide 4x^3 + 2x^2 + 10x - 7 by (x-5).
divide 4x^3 + 2x^2 + 10x - 7 by (x-5) using the long division..

this is polynomial method i think :/.. help please me.. :((Crying) i thought you will understand me.. :( thanks.. need the answer right now because i will pass this tommorow.. thanks :)
• Oct 18th 2010, 04:13 AM
Prove It

$\displaystyle \displaystyle{\frac{4x^3 + 2x^2 + 10x - 7}{x-5}}$?
• Oct 18th 2010, 05:54 AM
HallsofIvy
A simple way to write "x raised to the 3 power" is x^3.

In any case, x divides into 4x^3 4x^2 times. Multiplying all of x- 5 by 4x^2 and subtracting, 4x^3+ 2x^2+ 10x- 7- (4x^2- 20x^2)= 22x^2+ 10x- 7.

x divides into 22x^2 22x times. Multiplying all of x- 5 by 22x and subtracting, 22x^2+ 10x- 7- (22x^2- 110x)= 120x- 7.

x divides into 120x 120 times. Multiplying all of x- 5 by 120 and subtracting, 120x- 7- (120x- 600)= 593.

x- 5 divides into 4x^3+ 2x^2+ 10x- 7 gives a quotient of 4x^2+ 22x+ 120 with a remainder of 593.
• Oct 18th 2010, 04:44 PM
skeeter
primo candidate for synthetic division ...

Code:

5].......4.......2.......10.......-7 .................20......110.....600 --------------------------------------- .........4.......22......120.....593
quotient ... $\displaystyle 4x^2 + 22x + 120 + \frac{593}{x-5}$
• Oct 18th 2010, 06:49 PM
TheCoffeeMachine
Or let:

$\displaystyle \displaystyle \frac{4x^3 + 2x^2 + 10x - 7}{x-5} = (ax^2+bx+c)+\frac{r}{(x-5)}$

$\displaystyle \displaystyle \Rightarrow 4x^3 + 2x^2 + 10x - 7 = (x-5)(ax^2+bx+c)+r$

$\displaystyle \displaystyle \Rightarrow 4x^3 + 2x^2 + 10x - 7 = ax^3+(b-5 a)x^2+(c-5 b)x-5c+r$.

Comparing the coefficients:

$\displaystyle \displaystyle a = 4$, $\displaystyle b-5a = 2 \Rightarrow b = 22$, $\displaystyle c-5b = 10 \Rightarrow c = 120$

and finally the remainder $\displaystyle -5c+r = -7 \Rightarrow r = 593$.

So $\displaystyle \displaystyle \frac{4x^3 + 2x^2 + 10x - 7}{x-5} = (4x^2+22x+120)+\frac{593}{x-5}$
• Oct 19th 2010, 08:39 AM
Soroban
Hello, ejaykasai!

Quote:

$\displaystyle \text{Divide }4x^3 + 2x^2 + 10x - 7\text{ by }(x-5)\text{ using long division.}$

You want Long Division? . . . Here it is . . .

. . $\displaystyle \begin{array}{cccccccccc} &&&& 4x^2 &+& 22x &+& 120 \\ && -- & -- & -- & -- & -- & -- & -- \\ x-5 & | & 4x^3 &+& \;\;2x^2 &+& 10x &-& 7 \\ && 4x^3 &=& 20x^2 \\ && -- & -- & -- \\ &&&& 22x^2 &+& \;\;10x \\ &&&& 22x^2 &-& 110x \\ &&&& -- & -- & -- \\ &&&&&& 120x &-& \;\;\;7 \\ &&&&&& 120x &-& 600 \\ &&&&&& -- & -- & -- \\ &&&&&&&& 593 \end{array}$