4x^2 - 10x - 4 is divided by x-5
Answers
Answer:
Divide 4x4 + 1 + 3x3 + 2x by x2 + x + 2
First, I'll rearrange the dividend, so the terms are written in the usual order:
4x4 + 3x3 + 2x + 1
I notice that there's no x2 term in the dividend, so I'll create one by adding a 0x2 term to the dividend (inside the division symbol) to make space for my work.
long-division set-up: 4x^4 + 3x^3 + 0x^2 + 2x + 1 divided by x^2 + x + 2
Then I'll do the division in the usual manner. Dividing the 4x4 by x2, I get 4x2, which I put on top. Then I multiply through, and so forth, leading to a new bottom line:
(4x^4)/(x^2) = 4x^2; (4x^2)(x^2 + x + 2) = 4x^4 + 4x^3 + 8x^2; (4x^4 + 3x^3 + 0x^2 + 2x + 1) – (4x^4 + 4x^3 + 8x^2) = –x^3 – 8x^2 + 2x + 1, which is my new bottom line
Dividing –x3 by x2, I get –x, which I put on top. Then I multiply through, etc, etc:
(–x^3)/(x^2) = –x; (–x)(x^2 + x + 2) = –x^3 – x^2 – 2x; (–x^3 – 8x^2 + 2x + 1) – (–x^3 – x^2 – 2x) = –7x^2 + 4x + 1, which is my new bottom line
Dividing –7x2 by x2, I get –7, which I put on top. Then I multiply through, etc, etc:
(–7x^2)/(x^2) = –7; (–7)(x^2 + x + 2) = –7x^2 – 7x – 14; (–7x^2 + 4x + 1) – (–7x^2 – 7x – 14) = 11x + 15, which is my new bottom line
And then I'm done dividing, because the remainder is linear (11x + 15) while the divisor is quadratic. The quadratic can't divide into the linear polynomial, so I've gone as far as I can.
Then my answer is:
\mathbf{\color{purple}{4\mathit{x}^2 - \mathit{x} - 7 + \dfrac{11\mathit{x} + 15}{\mathit{x}^2 + \mathit{x} + 2}}}4x
2
−x−7+
x
2
+x+2
11x+15