Question

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

Answer 1

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:

$30 -29x - 7x^2+5x^3 + x^4=0$

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:

$30 -29x - 7x^2+5x^3 + x^4=0$

$30 -29*(1) - 7*(1)^2+5*(1)^3 + (1)^4=0$

$30 -29 - 7+5 + 1=0$

$0=0$ notice that x=1 is a root.

Let's try now with x=2, we have:

$30 -29x - 7x^2+5x^3 + x^4=0$

$30 -29*(2) - 7*(2)^2+5*(2)^3 + (2)^4=0$

$30 -58 - 28+40 + 16=0$

$0=0$ notice that x=2 is also a root.

So the original equation can be written as:

$30 -29x - 7x^2+5x^3 + x^4=(x-2)*(x-1)*p(x)$ so:

$(30 -29x - 7x^2+5x^3 + x^4)/((x-2)*(x-1))=p(x)$

$(30 -29x - 7x^2+5x^3 + x^4)/(x^2 - 3 x + 2)=p(x)$

Doing the math, we have:

$(30 -29x - 7x^2+5x^3 + x^4)-(x^2)*(x^2 - 3 x + 2)=$

$(30 -29x - 7x^2+5x^3 + x^4)-(x^4 - 3 x^3 + 2x^2)=$

$(30 -29x - 9x^2+8x^3)$ now:

$(30 -29x - 9x^2+8x^3)-(8x)*(x^2 - 3 x + 2)=$

$(30 -29x - 9x^2+8x^3)-(8x^3 - 24 x^2 + 16x)=$

$(30 -45x +15x^2)$ now:

$(30 -45x +15x^2)-(15)*(x^2 - 3 x + 2)=$

$(30 -45x +15x^2)-(15x^2 - 45 x + 30)=0$

which means:

$(30 -29x - 7x^2+5x^3 + x^4)/((x-2)*(x-1))=p(x)$

$(30 -29x - 7x^2+5x^3 + x^4)/((x-2)*(x-1))=(x^2+8x+15)$

Using the equation for calculating the roots of a quadratic equation we have:

quadratic equation: $x^2+8x+15$

$x=\frac{-b\±\sqrt{b^2-4ac} }{2a}$

$x=\frac{-8\±\sqrt{(8)^2-(4*1*15)} }{2*1}$

$x=\frac{-8\±\sqrt{(8)^2-(4*1*15)} }{2*1}$

$x=\frac{-8\±\2}{2}$

$x1=-5$

$x2=-3$

So finally we have:

$30 -29x - 7x^2+5x^3 + x^4=(x-2)*(x-1)*(x+5)*(x+3)$

In conclusion the roots are: x=2, x=1, x=-5 and x=-3.