Please Help Me!

Question

Please Help Me!

jacksonjrb · Answer

$$x^4-\frac{ 13 }{ 2 }x^3+8x^2-\frac{ 5 }{ 2 }x=0$$

jacksonjrb · Answer

Given one or mote of its roots, complete the solution set for the equation.
Need two more roots. Roots already given: (0, 5)

jacksonjrb · Answer

*more

mathstudent55 · Answer

First, factor out x.
What do you get?

jacksonjrb · Answer

$$x(x^3-\frac{ 13 }{ 2 }x^2+8x-\frac{ 5 }{ 2 })$$

jacksonjrb · Answer

=0

mathstudent55 · Answer

Of course, that is equal to zero.
From the zero product rule, you can see that x = 0 is a solution, which you already have.

jacksonjrb · Answer

Yes.

mathstudent55 · Answer

Now since you know that 5 is a solution, that means that that 3rd degree polynomial is divisible by x - 5

jacksonjrb · Answer

Okay.

mathstudent55 · Answer

You can use long division or synthetic division to divide by x - 5 to get a quadratic polynomial. Then you can try factoring the quadratic or use the quadratic formula.

jacksonjrb · Answer

Okay, that makes sense/

jacksonjrb · Answer

I'll try it, thanks.

mathstudent55 · Answer

Just one suggestion before you try.

jacksonjrb · Answer

Yes?

mathstudent55 · Answer

$x(x^3-\dfrac{ 13 }{ 2 }x^2+8x-\dfrac{ 5 }{ 2 })= 0$
Multiply both sides of the equation by 2 to eliminate the denominators.
Since the right side is zero anyway, it'll remain zero.
Then you get this:

mathstudent55 · Answer

$x \times \left[2 \times (x^3-\dfrac{ 13 }{ 2 }x^2+8x-\dfrac{ 5 }{ 2 }) \right]= 2 \times 0$

$x (2x^3-13x^2+16x-5)= 0$

mathstudent55 · Answer

Now divide $2x^3 - 13x^2 + 16x + 5$ by $x - 5$.

jacksonjrb · Answer

Okay, thank you.

mathstudent55 · Answer

You're welcome.