Solution to quadratic and cubic equation with partial root

Question

I am having trouble understanding how to resolve quadratic and cubic equations using the method described by my university lecturer (I am really interested to know if this method has a name). My calculator gives very different answers to what my lecturer explained and this enquiring mind wants to know why. The question was $$x^3+8x^2+16x+3=0$$ A given root was $x=-3$

While my lecturer gives the roots $x_{1,2,3}=1,5,1$. I'm not fully confident nor conversant with the methodology he employed yet. The general formula need not be related but the steps were something like the following: $$x^3+8x^2+16x+3=0$$ Where $0=(x+3)(ax^2+bx+c)$ we substitute the original formula values so, expanding the equation to $$ x^3+8x^2+16x+3=(x+3)(ax^2+bx+c) $$ And hence expansion to $$ =>ax^3+3ax^2+bx^2+3bx+cx+3c $$

Since the first factor is given as $(x+3)$ we need only find the others using any other factorisation method for quadratic equations.

This is where I got very lost. $$=ax^3+(3a+b)x^2+(3b+c)x+3c$$ By inspection a=1 and c=1 calculated by $$a: ax^3=x^3$$ $$a: a=1$$

$$b: (3a+b)x^2=8x^2$$ $$b: 3(1)+b=8$$ $$b: b=8-3=5$$

$$c: 3=3c$$ $$c: 1=c$$

So the final factors he explained were $(x+3)(x-1)(x-5)=0$

Is this the correct final answer or is the calculator doing something not explained to me?

Edit: @Gnumbertester states that the roots found are the quadratic equation values of $(x+3)(x^2+5x+1)$ I how this image helps other understand why there are 3 roots.

Fortunately I passed my degree module 1st class! Thanks to all who contributed.

Answer 1

Your answer is incorrect and there is a much simpler way of finding the roots.

Since you are given a root of this cubic, you can use synthetic division to reduce it to a quadratic.

You may already know how to do synthetic division, however if you don't, you can see how to do it here.

Once you divide $x^3+8x^2+16x+3$ by the linear factor you are given, you are left with $x^2+5x+1$.

To find the other two roots, use the quadratic formula.

Addendum: If you are familiar with traditional, polynomial long division, that works too.