38 votes
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Factorization of quartic polynomial.

Note that $$(x^2 + x + 1)^2 = x^4 + 2x^3 + 3x^2 + 2x + 1$$which means that your polynomial is equal to $$ (x^2 + x + 1)^2 - 4 \\ = (x^2 + x + 1)^2 - 2^2\\ = (x^2 + x + 1 - 2)(x^2 + x + 1 + 2) $$
Arthur's user avatar
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32 votes
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How to factor a fourth degree polynomial

The only really general way of which I am aware is to guess at the form of the factorization. Since it is monic (the highest term has coefficient 1), you know that the factors should also be so. Thus, ...
The_Sympathizer's user avatar
27 votes
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Why can a quartic polynomial never have three real and one complex root?

If $z$ is a root of a real polynomial, say $p(z) =\sum_{j=0}^n r_jz^j = 0$, then $\overline{z}$ is also a root of $p$ as $p(\overline{z}) =\sum_{j=0}^n r_j\overline{z}^j = \sum_{j=0}^n r_j\overline{z^...
quid's user avatar
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26 votes
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Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .

Write $$x^4-6x+6=(x^2-1)^2+2\left(x-\frac32\right)^2+\frac12$$ and then it is clear that $x^4-6x+6>0$ for all $x\in\mathbb{R}$.
A. Goodier's user avatar
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25 votes

How to factor a fourth degree polynomial

Let $$f (x) = x^4 + 10 x^3 + 39 x^2 + 70 x + 50$$ Converting to a depressed quartic see here, we see that the $x$ term drops out as well $$f\left(x-\frac{5}{2}\right)=x^4+\frac{3 x^2}{2}+\frac{25}{...
Lozenges's user avatar
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24 votes

Factorization of quartic polynomial.

One may try to use the rational root theorem. But that is not possible in this case because there is no rational root. Another thing we can try (if we don't "see" anything) is write $$(x^2+ax+b)(x^...
Arnaldo's user avatar
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17 votes
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What are the roots of $x^4 + x^2 + 1$?

Two possible approaches: solve $y^2+y+1=0$ and then $x^2=y$, observe that $(x^4+x^2+1)(x^2-1)=x^6-1$.
Angina Seng's user avatar
17 votes
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Finding the sum of squares of roots of a quartic polynomial.

We have that $$(x-a)(x-b)(x-c)(x-d)=$$ $$=x^4-(a+b+c+d)x^3+(ab+ac+ad+bc+bd+cd)x^2\\-(abc+abd+acd+bcd)x+abcd$$ then by $S_1=a+b+c+d$ $S_2=ab+ac+ad+bc+bd+cd$ $S_3=abc+abd+acd+bcd$ $S_4=abcd$ $$a^2+b^...
user's user avatar
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16 votes

Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification

Substitute $x=y+1$, then $z=y+1/y$ to get $$y^4+5y^3+9y^2+5y+1=0$$ $$y^2((y+1/y)^2+5(y+1/y)+7)=0$$ $$z^2+5z+7=0$$ Then $z=\frac{-5\pm\sqrt{-3}}2$ and $x=\frac{z\pm\sqrt{z^2-4}}2+1$, where the four ...
Parcly Taxel's user avatar
16 votes

Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .

Still another way : If $x≤0$, then $x^4-6x≥0\thinspace .$ Therefore, $x>0\thinspace .$ Thus, using the AM-GM inequality you have : $$x^3+\frac 2x+\frac 2x+\frac 2x=6≥4\sqrt [4]{8}$$ A contradiction ...
lone student's user avatar
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14 votes
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How do I solve this equation; $x^4+4x-1=0$?

You may write it as $$ x^4+4x-1=(x^2+1)^2-2(x-1)^2 = (x^2+1)^2-(\sqrt{2}(x-1))^2$$ and factorize.
H. H. Rugh's user avatar
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14 votes

Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .

We can use for example $$x^4-6x+6 =(x^2-1)^2+2x^2-6x+5 >0$$ indeed $(x^2-1)^2\ge 0$ and $$36-4\cdot 2\cdot 5 =-4<0$$
user's user avatar
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13 votes
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How to solve $x(3x+3)(x+5)(2x+12)+576 = 0$?

It is equivalent to $x(x+1)(x+5)(x+6)+96 = 0$ Now $$(x^2+6x)(x^2+6x+5)+96=0$$ Let $t=x^2+6x$ and finish the job...
nonuser's user avatar
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12 votes
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How many real roots does $x^4 - 4x^3 + 4x^2 - 10$ have?

Hint: Your polynomial can be factored into two quadratics using difference of squares: $$x^2(x-2)^2-10=(x(x-2)+\sqrt{10})(x(x-2)-\sqrt{10}).$$ Can you take it from here?
J. W. Tanner's user avatar
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12 votes

Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification

We can look for a difference of squares factorization. Completing the square gives $$\left( x^2 + \frac{1}{2} x + c \right)^2 - \left( 2c + \frac{1}{4} \right) x^2 - (c + 2) x - (c^2 - 1)$$ and we ...
Qiaochu Yuan's user avatar
12 votes

Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .

Hint: $\,P(x+1) = x^4 + 4 x^3 + 6 x^2 - 2 x + 1 = x^2(x+2)^2 + x^2 + (x-1)^2\,$.
dxiv's user avatar
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12 votes

Show that the polynomial $P(x):=x^4-6x+6$ has no real roots .

This question attracted a huge echo, so let us write down a further decomposition of $f$ into a sum of squares: $$ x^4-6x+6=\left(x^2 -\frac 32\right)^2 + 3(x-1)^2 + \frac 34\ge \frac 34=0.75\ . $$ $\...
dan_fulea's user avatar
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11 votes
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Solving these two equations

We have $$1+\frac{1}{x^2+y^2}=\frac{12}{5x}$$ and $$1-\frac{1}{x^2+y^2}=\frac{4}{5y},$$ which gives $$\frac{6}{x}+\frac{2}{y}=5$$ or $$y=\frac{2x}{5x-6},$$ which after substitution to the first ...
Michael Rozenberg's user avatar
11 votes
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How can I prove that $p(x)=x^4+x+1$ doesn't have real roots?

Consider three cases . . . If $x\ge 0$ then $ x^4+x+1\ge 1 $.$\\[4pt]$ If $-1 < x < 0$ then $ x^4+x+1 > x^4 + (-1) + 1 > 0 $.$\\[4pt]$ If $x\le -1$ then $ x^4+x+1 \ge x^2+x+1 = \Bigr(x+\...
quasi's user avatar
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10 votes

Why can a quartic polynomial never have three real and one complex root?

Who told a quartic polynomial can't have 3 real roots.? It can have provided the coefficients are complex... If coefficients have to be real then you must see that to cancel the $i$ of one root there ...
Qwerty's user avatar
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10 votes

How to factor a fourth degree polynomial

Hint: Make the ansatz $$x^4+10x^3+39x^2+70x+50=(x^2+ax+b)(x^2+cx+d)$$ Expanding the right-hand side $$x^4+x^3(a+c)+x^2(b+d+ac)+x(bc+ad)+bd$$ And you will get $$a+c=10,b+d+ac=39,bc+ad=70,bd=50$$
Dr. Sonnhard Graubner's user avatar
9 votes

Find all four roots of quartic equation $x^4-x+1=0$

There are no real solutions because $x^4-x+1$ attains a positive minimum at $x=1/\sqrt[3]{4}$.
lhf's user avatar
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9 votes

How to solve $x^4-2x^3-x^2+2x+1=0$?

Another way to notice the factorization $$x^4-2x^3-x^2+2x+1=0$$ Since $x=0$ is not the root of the equation, divide by $x^2$ to get $$x^2 -2x-1 + \frac{2}{x} + \frac{1}{x^2} = 0$$ Rewrite it as $$x^2+\...
VIVID's user avatar
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8 votes

Need help solving $x^4-3x^3-11x^2+3x+10=0$

One can check that $1$ and $-1$ are both roots of this polynomial. After factoring out $x-1$ and $x+1$, you're left with a quadratic polynomial. As a general rule, the rational root theorem is a good ...
carmichael561's user avatar
8 votes
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Factoring $x^4-11x^2y^2+y^4$

We notice this looks a bit like $(x^2-y^2)^2$, so we write \begin{align*} x^4 - 11x^2y^2 + y^4 &= (x^4 - 2x^2y^2+y^4) - 9x^2y^2 \\ &= (x^2-y^2)^2 - (3xy)^2 \\ &= (x^2 + 3xy - y^2)(x^2 - ...
Caleb Stanford's user avatar
8 votes

Find all four roots of quartic equation $x^4-x+1=0$

Note that $x^4-x+1=0$ is a deeply-depressed quartic equation, which makes it manageable. In fact, it can be factorized as $$x^4-x+1= \left( x^2- ax+ \frac{a^3-1}{2a} \right) \left( x^2+ ax+ \frac{a^3+...
Quanto's user avatar
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8 votes

Possible "clever" ways to solve $x^4+x^3-2x+1=0$, with methodological justification

Since this quartic has no real roots, it has two pairs of complex conjugate roots, so it must factor into two conjugate quadratics: $$(x^2 + ax + b)(x^2 + \overline ax + \overline b) = x^4 + (a + \...
Anders Kaseorg's user avatar

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