Question

Solve 2m^3 + 5m^2 - 13m -5=0, given -1/2 is a root

Answer 1

EXPLANATION:

Given;

We are given the following polynomial equation;

$2m^3+5m^2-13m-5=0$

Required;

We are required to solve the polynomial given that -1/2 is a root.

Step-by-step solution;

We are already told that one of the roots of the equation is -1/2, that is;

$x=-\frac{1}{2}$

We can therefore divide the polynomial by this root to get the quotient which would be a quadratic equation. This is shown below;

Let us now go over the solution together.

Using the synthetic division method, we start by listing out the coefficients of each term in the polynomial and that is;

$\begin{gathered} Coefficients: \\ 2,5,-13,-5 \end{gathered}$

Next step, we take down the first coefficient, which is 2. Next we take the root (that is -1/2) and multiply this by the first coefficient.

That gives us,

$2\times-\frac{1}{2}=-1$

We write out the result directly under the next coefficient (that is 5) and add them together. That results in (5 - 1 = 4).

Next step, we multiply 4 by the root and we have;

$4\times-\frac{1}{2}=-2$

We write this result directly under the next coefficient (that is -13) and add them together. That gives us -15. Multiply this result by the root and we'll have;

$-15\times-\frac{1}{2}=7\frac{1}{2}$

Write this result directly under the next coefficient and add up and we'll have 2 1/2 (or 2.5).

Note that we now have the new coefficients as;

$\begin{gathered} 2,4,-15 \\ Remainder\text{ }\frac{5}{2} \end{gathered}$

Therefore, the quotient is;

$2m^2+4m-15\text{ }+\frac{\frac{5}{2}}{(m+\frac{1}{2})}=0$

Using the quadratic equation formula, we now have the other roots solved as follows;

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

Where the variables are;

$a=2,b=4,c=-15$

Inputing the variables into the quadratic formula above, will now give us;

$\begin{gathered} m_1=-\frac{\sqrt{34}}{2} \\ m_2=-1+\frac{\sqrt{34}}{2} \end{gathered}$

We now have the solution as;

ANSWER:

$m=-\frac{1}{2},-\frac{\sqrt{34}}{2},-1+\frac{\sqrt{34}}{2}$