I know factoring questions are a dime a dozen but I can't seem to get this one. $-2x^3+2x^2+32x+40$ Factor to obtain the following equation: $-2(x-5)(x+2)^2$ Do I have to use division (I'd prefer...

The way to factor a four-term polynomial like this is to apply Rational Root Theorem along with synthetic division or substitution to determine whether a rational root works for the polynomial or not.

Dec 3, 2017 · 1 I have the polynomial: $$2x^5-x^4+10x^3-5x^2+8x-4$$ and I know that the final result is: $$ (2x-1) (x^4+5x^2+4) = (2x-1) (x^2+1) (x^2+4)$$ But how would you do it step by step? I've …

Oct 10, 2020 · For quadrinomials what is the go to method? For example if you try to factor this polynomial x3 + 2x2 + x − 4 x 3 + 2 x 2 + x 4 You would end up with: x2(x + 2) + 1(x − 4) x 2 (x + 2) + …

Jul 30, 2015 · I have a quartic in $\\Bbb Z[x]$ with very large coefficients that I know splits into two quadratics in $\\Bbb Z[x]$. What is the best way to do find the quadratics?

Jul 28, 2010 · Could you change the title to "Is there a general formula for solving 4th degree polynomial equations" or "Is there a general formula for solving quartic equations?"

Feb 7, 2015 · Is there a polynomial function $P(x)$ with real coefficients that has an infinite number of roots? What about if $P(x)$ is the null polynomial, $P(x)=0$ for all x?

Feb 4, 2019 · Is there a way to factorise $x^n + 1$ I thought of doing it like this: $$x^n +1 = (x+1)(x^{n-1} - x^{n-2} + \\cdots - x + 1)$$ but can't seem to get anywhere using ...

In this method, the algorithm is performed as the multiplication modulo quadnomial which is obtained by the irreducible polynomial which has m degree and m terms.

4 days ago · Consequently, for any integer $k \ge 1$: $$ x^ {2^k} \equiv 1 \pmod 4 $$ $$ x^ {2^k} + 1 \equiv 2 \pmod 4 $$ This means that every factor in the product $P$ contributes exactly one factor of …