# Solution to quadratic and cubic equation with partial root

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 $$(x+3)(x^2+5x+1)$$

• What's a "partial root"? – Gnumbertester Jan 31 at 2:39
• I'm not a mathematician so you'll have to correct me where the terminology is invalid. – Rhodie Jan 31 at 2:40
• This is wrong. It's obvious that $x^3+8x^2+16x+3$ has no positive roots. (If you substitute a positive number for $x,$ each term is positive.) In fact, $x^3+8x^2+16x+3=(x+3)(x^2+5x+1)$ – saulspatz Jan 31 at 2:49

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.

• What is a linear factor? – Rhodie Jan 31 at 2:47
• Since you are given the root $x=-3$, you know that $x+3$ is a factor of the cubic. I just included linear because it is a linear term (degree of 1). – Gnumbertester Jan 31 at 2:48
• Above my pay grade buddy... – Rhodie Jan 31 at 2:52
• Linear isn't technically necessary for the understanding of this problem. The most important part is knowing that if you are given a factor of a cubic, you can find an expression (quadratic) by dividing the cubic by that factor. From there, all you need to do is factor the quadratic and you have all 3 roots of the original cubic. – Gnumbertester Jan 31 at 2:55
• You lost me at linear... – Rhodie Jan 31 at 3:15