Question 710429
{{{2x^3 +21x^2 +mx + q =0}}}
Let's say that a root is "r". Because of the geometric progression we are given, the other two roots could be expressed as 2r and 4r. We can write an equation for a polynomial with these roots:
{{{a(x-r)(x-2r)(x-4r) = 0}}}
We now multiply this out:
{{{a(x-r)(x^2-6xr+8r^2) = 0}}}
{{{a(x^3-6x^2r+8xr-x^2r+6xr^2-8r^3) = 0}}}
{{{ax^3-6ax^2r+8axr^2-ax^2r+6axr^2-8ar^3) = 0}}}
Adding like terms:
{{{ax^3-7ax^2r+14axr^2-8ar^3) = 0}}}
This is a general equation for a polynomial whose roots are r, 2r and 4r. Our equation:
{{{2x^3 +21x^2 +mx + q =0}}}
must fit this pattern. Now we find the a, m and q that make our equation fit the pattern of the general equation. We have a term of 2x^3 and the general form has only one {{{x^3}}} term, {{{ax^3}}}. So a must be 2. Replacing the "a" in the general form with 2 we get:
{{{(2)x^3-7(2)x^2r+14(2)xr^2-8(2)r^3) = 0}}}
which simplifies as follows:
{{{2x^3-14x^2r+28xr^2-16r^3) = 0}}}<br>
Now the modified general form has one {{{x^2}}} term, {{{-14x^2r}}}. This must match the {{{x^2}}} term of our equation, {{{21x^2}}}. Using this we can find that r must be -3/2 in order for {{{-14x^2r}}} to be equal to {{{21x^2}}}. Replacing the r in our modified general form with -3/2:
{{{2x^3-14x^2(-3/2)+28x(-3/2)^2-16(-3/2)^3) = 0}}}
Simplifying...
{{{2x^3-14x^2(-3/2)+28x(9/4)-16(-27/8) = 0}}}
{{{2x^3+21x^2+63x+54 = 0}}}<br>
We can now see that "m" must be 63 and "q" must be 54.<br>
The only thing left is the roots. We have already found that one root, r, is -3/2. The other two roots were 2r and 4r. Using -3/2 for we will find that the other roots are -3 and -6.