# solving the equation $x^4-5x^3+11x^2-13x+6=0$ by given condition

1.(a) solve the equation $x^4-5x^3+11x^2-13x+6=0$ , given that two of its roots $p$ & $q$ are connected by the relation $3p+2q=7$

(b) solve the equation $x^4-5x^3+11x^2-13x+6=0$ which has two roots whose difference is $1$

did I need to solve these problems by taking roots $x_1,x_2,x_3,x_4$ and using the relation between roots and coefficients and the given facts. it will be then a very lengthy process.is there any alternative short process

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Polynomials, such as these, you should always consider 1,-1,2,-2 as a chose of your roots because in most cases that I dealt with they are. If you plug in 1 as a root you see that it satisfies the condition. From there you can factor out a $(x-1)$ from the equation. –  diimension Nov 20 '12 at 5:42
diimension: I think (in general) one would try all the factors of the constant term, here 6, so also 3,-3,6,-6 besides 1,-1 and 2,-2. –  coffeemath Nov 20 '12 at 5:54

Let $$f(x) = x^4 - 5x^3 + 11x^2 - 13x + 6$$ We get that $$f(1) = 1 - 5 + 11 - 13 + 6 = 0$$ and $$f(2) = 16 - 5 \times 8 + 11 \times 4 - 13 \times 2 + 6 = 16 - 40 + 44 - 26 + 6 = 0$$ Hence, we have that $$f(x) = (x-1)(x-2)(x^2 + ax + b)$$ Plugging in $x=0$, we get that $$f(0) = 2b = 6 \implies b = 3.$$ Plugging in $x=3$, we get that $$f(3) = 2(9 +3a+3) = 81-5 \times 27 + 11 \times 9 - 13 \times 3 + 6$$ This gives us that $$6a + 24 = 12 \implies a = -2$$ Hence, we have that $$f(x) = (x-1)(x-2)(x^2 - 2x+3)$$ Hence, $$f(x) = 0 \implies x = 1 \text{ or } x= 2 \text{ or }x^2 -2x+3 = 0$$ $$x^2 -2x+3 = 0 \implies (x-1)^2 + 2 =0 \implies x = 1 \pm i\sqrt{2}$$ Hence, the roots are $$x=1,2,1 \pm i \sqrt{2}$$

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You can use the Rational root theorem to find rational roots and by dividing in $x-r$ where $r$ is a root you need to find the roots of a polynomial of smaller degree.

In a) I get that $r=1$ is a root, the division yields $x^3-4x^2+7x-6$ , now using the theorem again we see that $2$ is a root, dividing by $x-1$ yields a polynomial of second degree, which you know how to find roots for.

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I agree the rational root theorem is an easier approach than the one asked for, but that wasn't the assignment. –  Ross Millikan Nov 20 '12 at 5:49
@RossMillikan - I answered "is there any alternative short process" which I tooked to be the OP question –  Belgi Nov 20 '12 at 5:52

For 1) you have that $p=\frac 13(7-2q)$, so $(x-q)(x-\frac 13(7-2q))=\frac 13(x-q)(3x-7+2q)$ divides into your polynomial. Divide that in and let $q$ be a value that leaves zero remainder.

For 2) you have that $(x-p)(x-p-1)$ divides into your polynomial.

That said, the rational root theorem seems an easier approach.

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