How to find the roots of polynomial equation #2x^3+2x^2-19x+20=0#?

1 Answer
Mar 29, 2017

Answer:

Solution set is #{-4, (3 + i)/2, (3 - i)/2}#

Explanation:

NOTE: I WROTE THIS ANSWER THINKING THE QUESTION WAS TO SOLVE FOR #x#, NOT JUST TO FIND THE ROOTS

Step 1: List the possible factors of this equation

This will be done using the rational root theorem, which states that the factors of #f(x) = px^n + qx^m + ... + C# are given by #"Factors of C"/"Factors of p"#, where #p# is the coefficient of the highest power of #x#.

Therefore, our possible factors are given by #(+- 1, +-2, +-4, +-5, +-10, +-20)/(+-2, +-1)#

Step 2: Check these factors using the remainder theorem

This is when the trial and error process comes into play. The remainder theorem is useful because you can check the remainders without actually having to do the division.

It states that if #x - a# is a factor of #f(x)#, then #f(a) = 0#. With a little bit of trial and error, we find that #x = -4# is a root because if #f(x) = 2x^3 + 2x^2 - 19x + 20#, then #f(-4) = 2(-4)^3 + 2(-4)^2 - 19(-4) + 20 = 0#.

Step 3: Use long division to simplify the equation

We have now found one root. We're not quite certain of the others yet. We can now simplify the equation to a quadratic by using long division or synthetic division.

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According, to the table above, the quotient is #2x^2 - 6x + 5#. The equation becomes #(x + 4)(2x^2 - 6x + 5)#. We now solve for #x# in the quadratic using either the quadratic formula or completing the square.

#x = (-(-6) +- sqrt((-6)^2 - 4(2)(5)))/(2(2))#

#x = (6 +- sqrt(-4))/4#

#x = (6 +- 2sqrt(-1))/4#

We now use the identity #sqrt(-1) = i#.

#x = (6 +- 2i)/4#

#x = (3 +- i)/2#

Hopefully this helps!