Q:

Consider the given polynomial P(x)=30 29x - 7x^2+5x^3 + x* .4 List all possible rational roots, and factor the given polynomial

Accepted Solution

A:
Answer:x=2, x=1, x=-5 and x=-3.Step-by-step explanation:Let's find the answer using the following method.Original equation:[tex]30 -29x - 7x^2+5x^3 + x^4=0[/tex]Notice that the constant term is 30, which can be divided by:1, -1, 2, -2, 3, -3, 5, -5, 6, -6, 10, -10, 15, -15, 30, -30.Trying this divisors can allowed us to find one or even two initial roots, so:Trying first x=1, we have:[tex]30 -29x - 7x^2+5x^3 + x^4=0[/tex][tex]30 -29*(1) - 7*(1)^2+5*(1)^3 + (1)^4=0[/tex][tex]30 -29 - 7+5 + 1=0[/tex][tex]0=0[/tex] notice that x=1 is a root.Let's try now with x=2, we have:[tex]30 -29x - 7x^2+5x^3 + x^4=0[/tex][tex]30 -29*(2) - 7*(2)^2+5*(2)^3 + (2)^4=0[/tex][tex]30 -58 - 28+40 + 16=0[/tex][tex]0=0[/tex] notice that x=2 is also a root.So the original equation can be written as:[tex]30 -29x - 7x^2+5x^3 + x^4=(x-2)*(x-1)*p(x)[/tex] so:[tex](30 -29x - 7x^2+5x^3 + x^4)/((x-2)*(x-1))=p(x)[/tex] [tex](30 -29x - 7x^2+5x^3 + x^4)/(x^2 - 3 x + 2)=p(x)[/tex] Doing the math, we have:[tex](30 -29x - 7x^2+5x^3 + x^4)-(x^2)*(x^2 - 3 x + 2)=[/tex] [tex](30 -29x - 7x^2+5x^3 + x^4)-(x^4 - 3 x^3 + 2x^2)=[/tex] [tex](30 -29x - 9x^2+8x^3)[/tex] now:[tex](30 -29x - 9x^2+8x^3)-(8x)*(x^2 - 3 x + 2)=[/tex][tex](30 -29x - 9x^2+8x^3)-(8x^3 - 24 x^2 + 16x)=[/tex][tex](30 -45x +15x^2)[/tex] now:[tex](30 -45x +15x^2)-(15)*(x^2 - 3 x + 2)=[/tex][tex](30 -45x +15x^2)-(15x^2 - 45 x + 30)=0[/tex]which means: [tex](30 -29x - 7x^2+5x^3 + x^4)/((x-2)*(x-1))=p(x)[/tex] [tex](30 -29x - 7x^2+5x^3 + x^4)/((x-2)*(x-1))=(x^2+8x+15)[/tex] Using the equation for calculating the roots of a quadratic equation we have:quadratic equation: [tex]x^2+8x+15[/tex] [tex]x=\frac{-b\±\sqrt{b^2-4ac} }{2a}[/tex][tex]x=\frac{-8\±\sqrt{(8)^2-(4*1*15)} }{2*1}[/tex][tex]x=\frac{-8\±\sqrt{(8)^2-(4*1*15)} }{2*1}[/tex][tex]x=\frac{-8\±\2}{2}[/tex][tex]x1=-5[/tex][tex]x2=-3[/tex]So finally we have:[tex]30 -29x - 7x^2+5x^3 + x^4=(x-2)*(x-1)*(x+5)*(x+3)[/tex] In conclusion the roots are: x=2, x=1, x=-5 and x=-3.