How do you divide #(x^2+5x-3)div(x-10)# using long division?

Question

Theodore Amidon · Accepted Answer

#(x^2+5x-3)div(x-10) = (x+15) " rem "147#
OR
#(x^2+5x-3)div(x-10) = (x+15) +147/(x-10)#
OR
#(x^2+5x-3) = (x-10)(x+15) +147# Algebraic long division follows the same method as arithmetic long division... #("dividend")/("divisor") = "quotient"# Step 1. Write the dividend in the 'box' making sure that the indices are in descending powers of x. Step 2. Divide the first term in divisor into the term in the dividend with the highest index. Write the answer at the top, Step 3. Multiply by BOTH terms of the divisor at the side Step 4. Subtract Step 5. Bring down the next term Repeat steps 2 to 5 #color(white)(xxxxxxxxxxxx)color(red)(x )color(blue)( + 15)" rem " 147# #color(white)(xx)x-10 |bar( x^2 +5x -3)" "larr x^2divx = color(red)(x)# #color(white)(xxx.xx)ul(color(red)(-(x^2-10x)))color(white)(.)darr" "larr# subtract (change signs) #color(white)(xxxxxxxxxxx) 15x-3""larr# bring down the -3, #15x div x = color(blue)(15# #color(white)(xxxxx.xxx)ul(color(blue)(-(15x-150))" "larr# subtract (change signs) #color(white)(xxxxxxxxx.xxxxxx)147 " "larr#remainder #(x^2+5x-3)div(x-10) = (x+15) " rem "147# This can also be written as #(x^2+5x-3)div(x-10) = (x+15) +147/(x-10)# Or#" "(x^2+5x-3) = (x-10)(x+15) +147#

Sadie Alexander · Answer

undefined To divide $x^2 + 5x - 3$ by $x - 10$ using long division, follow these steps:

1. Divide the first term of the dividend by the first term of the divisor: $x^2$ divided by $x$ equals $x$.
2. Multiply the divisor by the result obtained in step 1: $x \times (x - 10)$ equals $x^2 - 10x$.
3. Subtract the result obtained in step 2 from the dividend: $x^2 + 5x - 3 - (x^2 - 10x)$ equals $15x - 3$.
4. Bring down the next term from the dividend: Bring down $15x$.
5. Repeat steps 1-3: Divide $15x$ by $x$, which equals $15$. Multiply $x - 10$ by $15$, resulting in $15x - 150$. Subtract this from $15x - 3$, which equals $147$.

So, $x^2 + 5x - 3$ divided by $x - 10$ equals $x + 15$ with a remainder of $147$.