divide: 6x⁵+4x⁴-3x³-1 by 3x²-x+1
Answers
Answer:
the answer is----- 6x^5+12x^4-9x^3-x^2-3x+3
hope it help you..
Answer:
2x^{3} +2x^{2} -x-12x
3
+2x
2
−x−1
Step-by-step explanation:
Rewrite
\begin{gathered}\\\frac{6x^{5}+4x^{4}-3x^{3} -1 }{3x^{2} -x+1}\end{gathered}
3x
2
−x+1
6x
5
+4x
4
−3x
3
−1
Factor the expressions
\frac{3x^{3}x(2x^{2} -1)+(2x^{2} -1)x(2x^{2} +1) }{3x^{2} -x+1}
3x
2
−x+1
3x
3
x(2x
2
−1)+(2x
2
−1)x(2x
2
+1)
\frac{3x^{3} (\sqrt{2} x- 1)x (\sqrt{2}x+1)+(\sqrt{2}x-1)x(\sqrt{2} x+1)x(2x^{2} +1) }{3x^{2} -x+1}
3x
2
−x+1
3x
3
(
2
x−1)x(
2
x+1)+(
2
x−1)x(
2
x+1)x(2x
2
+1)
\frac{(\sqrt{2}x-1)x(\sqrt{2} x+1)x(3x^{3}+2x^{2} +1) }{3x^{2} -x+1}
3x
2
−x+1
(
2
x−1)x(
2
x+1)x(3x
3
+2x
2
+1)
Rewrite the expression
\frac{(\sqrt{2}x-1)x(\sqrt{2}x+1)x(3x^{3} +3x^{2} -x^{2} +1) }{3x^{2} -x+1}
3x
2
−x+1
(
2
x−1)x(
2
x+1)x(3x
3
+3x
2
−x
2
+1)
Rewrite and reorder
\frac{(\sqrt{2}x-1)x(\sqrt{2}x+1)x(3x^{2} x(x+1)+1-x^{2} ) }{3x^{2} -x+1}
3x
2
−x+1
(
2
x−1)x(
2
x+1)x(3x
2
x(x+1)+1−x
2
)
Factor the expression
\frac{(\sqrt{2}x-1)x(\sqrt{2}x+1)x(3x^{2} x(x+1)+1-x^{2} }{3x^{2} -x+1}
3x
2
−x+1
(
2
x−1)x(
2
x+1)x(3x
2
x(x+1)+1−x
2
\frac{(\sqrt{2}x-1)x(\sqrt{2} x+1)x(x+1)x(3x^{2} +1-x) }{3x^{2} -x+13x 2−x+1(2
x−1)x(
2
x+1)x(x+1)x(3x
2
+1−x)
Reduce the fraction
(\sqrt{2}x-1)x(\sqrt{2} x+1)x(x+1)(
2
x−1)x(
2
x+1)x(x+1)
Simplify the product
(2x^{2} -1)x(x+1)(2x
2
−1)x(x+1)
Remove the parentheses
2x^{3} +2x^{2} -x-12x
3
+2x
2
−x−1