Question

Find the roots of the polynomial equation.

2x3 + 2x2 – 19x + 20 = 0

I seriously don't understand this at all

Answer 1

ok.. we start with the "rational root test" to find the first one

Answer 2

p: factors of constant term 20
\(\pm1,\pm2,\pm4,\pm5,\pm10,\pm20\)

q: factors of the coefficient of the highest power of "x" = 2
\(\pm1,\pm2\)

Answer 3

then, by the rational-root theorem,
if the given polynomial has a rational root, it must be one of the possible (p/q)\[
\pm{1\over1}=1,\pm{2\over1}=2,\pm{4\over1}=4,\ldots
\]

Answer 4

follow?

Answer 5

yes

Answer 6

evaluate

Answer 7

ok, then by remainder theorm, we check each possibility one -by-one
1) x=1
\[2(1)^3+2(1)^2-19(1)+20\ne0\]
so, x=1 is not a root
2) x=-1
\[2(-1)^3+2(-1)^2-19(-1)+20\ne0\]
again, x=-1 is not a root

Answer 8

3) x=2
\[2(2)^3+2(2)^2-19(2)+20\ne0\]
not a root

4)x=-2
\[2(-2)^3+2(-2)^2-19(-2)+20\ne0\]
not a root!!

Answer 9

5) x=4
\[2(4)^3+2(4)^2-19(4)+20\ne0\]
not a root

6)x=-4
\[2(-4)^3+2(-4)^2-19(-4)+20=\mathbf{0}\]
Houston,m we have a winner

Answer 10

we found one rational root.
\(\Large{x = -4}\)

for the next step, did you study the 'Descartes Rule of Signs'??

Answer 11

don't know what that is

Answer 12

not a problem..
but kapeesh so far?

Answer 13

yes

Answer 14

good.
Now, since the polynomial is of degree "3", there are '3' roots for it.
and we found one
for the other two, we have to take out what we found already..
so,
\[(x+4)(\ldots)=2x^3+2x^2-19x+20\]

Answer 15

the term in the paranthesis will be a quadratic (3-1=2) which we can get using polynomial division or sythethetic division

Answer 16

you with me?