Answer:
[tex]x=1, x=3, x=7[/tex]
Step-by-step explanation:
Start multiplying and adding all together. You get[tex]x^3-11x^2+31x-21 =0[/tex].
At this point we're stuck factoring since there exist a formula for the roots, but it's nowhere easy to remember. We start writing down The List (rational root theorem: it's the divisor of the constant term divided by the divisor of the lead coefficent, with a plus or minus, in our case it's [tex]\pm1, \pm3, \pm7, \pm 21[/tex]) and let's try checking one by one.
Value [tex]+1[/tex] gives a zero on the LHS, it means the polynomial is divisible by [tex](x-1)[/tex]. By doing the division, or writing it as [tex](x-1)(ax^2+bx+c) = x^3-11x^2+31x-21[/tex], multiplying on the LHS and then finding the values of a, b, and c (easiest for me, your mileage may vary) we can write our expression, finally, as
[tex](x-1)(x^2-10x+21) =0[/tex] At this point you can either apply the quadratic formula or decompose again the second factor as
[tex](x-3)(x-7)[/tex] by finding two numbers whose sum is -10 and product is 21, ie -3 and -7. At this point the expression
[tex](x-1)(x-3)(x-7)=0[/tex] is true if [tex]x=1, x=3, x=7[/tex]
