# How do you find the real solutions of the polynomial #2x^3=19x^2-49x+20#?

##### 1 Answer

The roots are

#### Explanation:

Given:

#2x^3=19x^2-49x+20#

Subtract the right hand side from the left to get the standard form:

#2x^3-19x^2+49x-20 = 0#

By the rational root theorem, any rational zeros of this cubic are expressible in the form

That means that the only possible *rational* roots are:

#+-1/2, +-1, +-2, +-5/2, +-4, +-5, +-10, +-20#

We find that

#2(color(blue)(1/2))^3-19(color(blue)(1/2))^2-49(color(blue)(1/2))+20 = (1-19-98+80)/4 = 0#

Hence

#2x^3-19x^2+49x-20 = (2x-1)(x^2-9x+20)#

To factor the remaining quadratic, find a pair of factors of

#x^2-9x+20 = (x-5)(x-4)#

So the other two roots are