Question

Simplify $\frac{5x}{x^2 - x - 6} + \frac{4}{x^2 + 4x + 4}$

score 133 · Answer 1 · 2026-03-31 13:55:46

133

Views

Simplify $\frac{5x}{x^2 - x - 6} + \frac{4}{x^2 + 4x + 4}$

Published on 31 Mar 2026 - 13:55

score 83 · Answer 2 · 2016-04-13 23:36:17

83

Views

finding factors

Published on 13 Apr 2016 - 23:36

score 124 · Answer 3 · 2016-04-14 08:44:27

124

Views

What will be the $(b^2-a)$?

Published on 14 Apr 2016 - 8:44

score 37 · Answer 4 · 2016-04-18 13:29:21

37

Views

Simplifying with exponents

Published on 18 Apr 2016 - 13:29

score 51 · Answer 5 · 2016-04-20 13:36:52

51

Views

What is the name of this kind of factoring algorithm

Published on 20 Apr 2016 - 13:36

score 28 · Answer 6 · 2016-04-20 15:45:27

28

Views

What is the theory of finding roots of a polynomial equation by looking at the factors of the $a_n$ and $a_0$ term called?

Published on 20 Apr 2016 - 15:45

score 1.9k · Answer 7 · 2016-04-20 16:56:20

1.9k

Views

Find the cubic polynomial given linear reminders after division by quadratic polynomials?

Published on 20 Apr 2016 - 16:56

score 48 · Answer 8 · 2016-04-26 16:06:17

48

Views

Find A and B for $A(x^{2}+3)+B(x^{3}+2)=5x^{4}+9x^{2}+4x$

Published on 26 Apr 2016 - 16:06

score 11.8k · Answer 9 · 2026-03-27 01:13:26

11.8k

Views

Factoring a 4th degree trinomial

Published on 27 Mar 2026 - 1:13

score 150 · Answer 10 · 2026-04-08 05:15:36

150

Views

What's the best way to compute $\frac{a^4 + b^4 + c^4}{a^2 + b^2 + c^2}$

Published on 08 Apr 2026 - 5:15

score 844 · Answer 11 · 2026-04-01 21:52:11

844

Views

Use Fermat factorization to factor $809009\ldots$

Published on 01 Apr 2026 - 21:52

score 1.2k · Answer 12 · 2016-05-06 18:22:16

1.2k

Views

finding the number of factors $2^{15}\times3^{10}\times5^6$

Published on 06 May 2016 - 18:22

score 512 · Answer 13 · 2016-05-07 11:37:40

512

Views

Highschool Exam Question About Cube Factoring

Published on 07 May 2016 - 11:37

score 63 · Answer 14 · 2016-05-07 12:46:33

63

Views

Number of solutions for $n^5 + 2 n^4 + n^3 - 3n + 2 $ mod $ 23^2 = 0$, where $0 \leq n < 23^2$ and $n\in \mathbb{N}$

Published on 07 May 2016 - 12:46

score 1.1k · Answer 15 · 2016-05-08 05:16:31

1.1k

Views

Factorization of Taylor series.

Published on 08 May 2016 - 5:16

List Question

Simplify $\frac{5x}{x^2 - x - 6} + \frac{4}{x^2 + 4x + 4}$

finding factors

What will be the $(b^2-a)$?

Simplifying with exponents

What is the name of this kind of factoring algorithm

What is the theory of finding roots of a polynomial equation by looking at the factors of the $a_n$ and $a_0$ term called?

Find the cubic polynomial given linear reminders after division by quadratic polynomials?

Find A and B for $A(x^{2}+3)+B(x^{3}+2)=5x^{4}+9x^{2}+4x$

Factoring a 4th degree trinomial

What's the best way to compute $\frac{a^4 + b^4 + c^4}{a^2 + b^2 + c^2}$

Use Fermat factorization to factor $809009\ldots$

finding the number of factors $2^{15}\times3^{10}\times5^6$

Highschool Exam Question About Cube Factoring

Number of solutions for $n^5 + 2 n^4 + n^3 - 3n + 2 $ mod $ 23^2 = 0$, where $0 \leq n < 23^2$ and $n\in \mathbb{N}$

Factorization of Taylor series.

Trending Questions

Popular # Hahtags

Popular Questions