12
votes
Accepted
Is there a method to factor this equation?
This is just factoring out the common factor of $x^2$ from the first two terms, and the common factor of $-4$ from the last two terms. However, the correct factorization would be
$$x^2(x+1) - 4(x+1)$$
...
- 36.7k
3
votes
Accepted
How to tell if a polynomial is square
The coefficients of $x_i^4$ in $p$ will be the squares of the coefficients of $x_i^2$ in $q$. Because of signs, there are $2^6=64$ very similar possibilities.
Then use the coefficients of $x_i^3x_j$ ...
- 48.7k
2
votes
Is there an easy way to tell whether a cubic or quartic polynomial is factorable over the integers?
There isn't a one-size-fits-all solution. There are strategies that one could try, for example reduction modulo a prime or a prime power. There are some well known criteria for detecting whether a ...
- 2,587
2
votes
Accepted
Terminology. Can I say that $t^2 - 9$ is factorable by a "product of conjugates"?
The answer to whether or not terminology is appropriate is very often; depends. If I had to give a direct answer, I would say no. Here is my reasoning.
In number theory, your terminology at least has ...
- 1,278
2
votes
Is there a method to factor this equation?
Observe that for $p(x) = x^3+x^2 - 4x - 4$, then $p(-1) = 0$. Thus as a polynomial, $p(x)$ has a factor $x+1$. You can then use synthetic division to obtain the quotient and can factor $p(x)$ into a ...
- 5,085
1
vote
Prove that the polynomial $f(x)= x^4 + 2x^2 + 2x + 1998$ cannot be writen as the product of two quadratic polynomials with integer coefficients
Easiest way: use a calculator like Wolfram Alpha to compute the four roots, then check every possible product $(x-\alpha)(x-\beta)$, where $\alpha$ and $\beta$ are roots.
- 6,361
1
vote
Accepted
Prove that the polynomial $f(x)= x^4 + 2x^2 + 2x + 1998$ cannot be writen as the product of two quadratic polynomials with integer coefficients
Use only $a+c=o, ad+bc=2\Leftrightarrow a(d-b)=2\Rightarrow d=b\pm1, \pm2$ while bd=1998. Then try to solve the resulting quadratic in b.
- 1,918
1
vote
Accepted
How has this been simplified?
In the above expression, you have a confusing expression:
$$36x\left(9x^2+3\right)\left(2x-8\right)-2\left(9x^2+3\right)^2$$
One way to see what happened is to simplify it by letting $y=9x^2+3$, and ...
- 6,094
1
vote
How has this been simplified?
It's just a matter of arithmetic:
$$\frac{36x(9x^2+3)(2x-8)-2(9x^2+3)^2}{(2x-8)^2}=\frac{ 18x\cdot \color{#66F}{2(9x^2+3)}(2x-8)-\color{#66F} {2(9x^2+3)}(9x^2+3)}{(2x-8)^2}= $$
$$=\frac{ 2(9x^2+3)[18x(...
- 2,731
1
vote
Is there a method to factor this equation?
$-4(x+1)$, you mean. You then factor out the $(x+1)$, obtaining $(x^2-4)(x+1)$. This is often called factoring by grouping.
- 107k
1
vote
How can I prove every integer of the form $6^{3k} + 1$ is composite, where k is a positive integer?
Hint
Remember that $a^3+b^3=(a+b)(a^2-ab+b^2)$. Now call $a=6^k$ and $b=1$.
Can you finish?
- 21.2k
1
vote
How can I prove every integer of the form $6^{3k} + 1$ is composite, where k is a positive integer?
You can factorise it as suggested by the comments (which is probably the standard way to answer the question), but you had a very good idea.
Assuming $3k$ is odd, as you noted, we can write:
$$6^{3k} -...
- 1,412
1
vote
Accepted
Two combinatorics questions, one on product of combinations, the other on why factoring heuristic isn't applicable
In a group of $14$ students, there are $8$ girls and $6$ boys. Determine the number of ways that a committee of $4$ students which has at least $1$ boy can be chosen from the group.
Let's examine two ...
- 72k
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