Divide 3x^3 + 4 by x + 1 and check.

Answer 1

-1 | 3 0 0 4

-3 3 -3

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3 -3 3 1

the quotient is 3x^2 - 3x + 3 with a remainder of 1

(x+1)(3x^2 - 3x + 3) + 1 = 3x^3 - 3x^2 + 3x

+3x^2 - 3x +3 + 1

= 3x^3 + 4

Answer 2

This idea of dividing is to see "what times the divisor is the dividend" In this case x+1 is the divisor. We'll go through seeing what we have to multiply our divisor by to get the largest term in our dividend (3x³ + 4). Once you get the largest term, you move down the line until you can't go any more.

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(x+1) | 3x³ + 4

x can go into 3x³ 3x² times (because x¹ times 3x² becomes 3x³

we then have to subtract 3x² * (x+1) which is 3x³ + 3x²

3x²

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(x+1) | 3x³ + 0x² + 0x + 4

-(3x² + 3x²⁾

-3x² + 0x + 4

The next term we have to worry about is -3x². The question is, what times x is -3x². The answer is -3x. So we subtract -3x * (x+1) which is -3x² - 3x

3x² - 3x

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(x+1) | 3x³ + 0x² + 0x + 4

-(3x² + 3x²⁾

-3x² + 0x + 4

-(-3x² - 3x)

3x + 4

Our last term is 3x. What times x is 3x? The answer is 3. So we'll have to subtract 3*(x+1) which is 3x+3

3x² - 3x + 3

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(x+1) | 3x³ + 0x² + 0x + 4

-(3x² + 3x²⁾

-3x² + 0x + 4

-(-3x² - 3x)

3x + 4

-(3x + 3)

1

We have that last one, but we shouldn't continue and do 1/x * x is 1 because if x is zero we'll have a problem. Therefore we just write the last bit over our divisor (x+1). Therefore, our final answer is...

3x² - 3x + 3 + ( 1/ (x+1) )

To check our work we can multiply our answer with our divisor (x+1)

(3x² - 3x + 3 + ( 1/ (x+1) )) * (x+1)

Im going to write each term times x, then each term times 1.

3x³ - 3x² + 3x + ( x/ (x+1) ) + 3x² - 3x + 3 + ( 1/ (x+1) )

I'll write is so you can see common terms in the same column

3x³ - 3x² + 3x + ( x/ (x+1) )

+ 3x² - 3x + 3 + ( 1/ (x+1) )

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3x³ + 0 + 0 + 3 + x/(x+1) + 1/(x+1)

3x³ + 3 + (x+1)/(x+1)

3x³ + 3 + 1

3x³ + 4