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
 A: 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 $$x=1,2,1 \pm i \sqrt{2}$$
A: 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.
A: 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.
A: $$x^4-5x^3+11x^2-13x+6=0$$
$$x^4-x^3-4x^3+4x^2+7x^2-7x-6x+6=0$$
$$(x-1)x^3-(x-1)4x^2+(x-1)7x-(x-1)6=0$$
$$(x-1)(x^3-4x^2+7x-6)=0$$
$$(x-1)(x^3-2x^2-2x^2+4x+3x-6)=0$$
$$(x-1)\left( (x-2)x^2-(x-2)2x+(x-2)3 \right)=0$$
$$(x-1)(x-2)(x^2+2x+3)=0$$
