I saw that "once have a root, we can divide the polynomials" and somehow
$\frac{3n^3 - 39n^2 + 360n + 20}{3(n-n_1)} = n^2 + (n_1 - 13)n + (n_1^2 - 13n_1 + 120)$ I'm noticing a pattern that there's no $n_1$, then have $n_1$, then $n_1^2$ not sure what rule is being used, if any. I'm not sure how to get the expression on the right hand side of the equals sign. This is a question I have for part of the answer given here
$\endgroup$ 12 Answers
$\begingroup$Let us make it using another way.
What you want is basically to find the coefficients $(a,b,c)$ such that$$3n^3 - 39n^2+360 n + 20 =3(n-n_1)(a n^2+b n+c)$$Expand and group terms to get$$(3 c n_1+20)+n (3 b n_1-3 c+360)+n^2 (3 a n_1-3 b-39)+(3-3 a) n^3=0$$ Now, cancel each coefficient. This gives $\color{red}{a=1}$ (that we could have guessed before starting); then remains$$(3 c n_1+20)+n (3 b n_1-3 c+360)+n^2 (-3 b+3 n_1-39)=0$$ giving $\color{red}{b=n_1-13}$; then remains$$(3 c n_1+20)+n (-3 c+3 (n_1-13) n_1+360)=0$$ from which $\color{red}{c=n_1^2-13 n_1+120}$; then remains$$3 n_1^3-39 n_1^2+360 n_1+20=0$$ which is true since $n_1$ is the first compute root of the equation.
Is this clear ?
$\endgroup$ 1 $\begingroup$It's just synthetic division. If we assume $n_1$ is a root and that $3n_1^3 - 39n_1^2 + 360n_1 + 20 = 0$
Then we can manipulate:
$3n^3 - 39n^2+360 n + 20 = $
$3(n^3 - 13n^2 + 120n + \frac {20}3) = $
$3(n^2*n - 13n^2 + 120n + \frac {20}3) = $
$3(n^2*(n-n_1) + n_1*n^2 - 13n^2 + 120n + \frac {20}3) = $
$3(n^2*(n-n_1) + (n_1 -13)n^2 + 120n + \frac {20}3) =$
$3(n^2*(n-n_1) + (n_1 - 13)n*n + 120n + \frac {20}3) =$
$3(n^2*(n-n_1) + (n_1 -13)n(n-n_1) + n_1(n_1-13)n + 120n + \frac {20}3)=$
$3(n^2*(n-n_1) + (n*n_1 - 13n)(n-n_1) + (n_1^2-13n_1 + 120)n + \frac {20}3)=$
$3(n^2*(n-n_1) + (n*n_1 - 13n)(n-n_1) + (n_1^2-13n_1 + 120)(n-n_1) + (n_1^2-13n_1 + 120)n_1 + \frac {20}3) =$
$3(n^2*(n-n_1) + (n*n_1-13n) (n-n_1) + (n_1^2 - 13n_1 + 120)(n-n_1) + \frac {3n_1^3-39n_1^2 + 360n_1 + 20}3)=$
$[3(n-n_1)](n^2 + (n_1-13)n + (n_1^2 -13n_1 + 120)) + [3n_1^3-39n_1^2 + 360n_1 + 20]=$
$[3(n-n_1)](n^2 + (n_1-13)n + (n_1^2 -13n_1 + 120)) + 0 =$
$[3(n-n_1)](n^2 + (n_1-13)n + (n_1^2 -13n_1 + 120))$
==== okay column notation===
$3n^3 - 39n^2 + 360n + 20\div 3 = n^3 - 13n^2 + 120n +\frac {20}3$.
And $(n^3 - 13n^2 + 120n +\frac {20}3)\div (n-n_1)=$
$(n-n_1)\begin{cases}n^2&+(n_1-13)n&+(n_1^2-13n_1+120)\\===&===&===&===\\n^3 & -13n^2 & +120n & \frac {20}3 \\n^2*n&-n_1*n^2\\----&----\\&(n_1-13)n^2&+120n\\&(n_1-13)n*n&-(n_1-13)n*n_1\\&----&----\\&&(n_1^2-13n_1+120)n&\frac {20}3\\&&(n_1^2-13n_1+120)n&-(n_1^2-13n_1+120)n_1\\&&----&----\\&&&(n_1^3 -13n_1^2+120n_1 + \frac {20}3) \end{cases}$
So $ {n^3 - 13n^2 + 120n +\frac {20}3} = ({n-n_1})(n^2+(n_1-13)n+(n_1^2-13n_1+120))+ ({n_1^3 -13n_1^2+120n_1 + \frac {20}3})$ and
$3n^3 - 39n^2 + 360n +20 = 3(n-n_1)(n^2+(n_1-13)n+(n_1^2-13n_1+120))+(3n_1^3 -39n_1^2+360n_1 + 20)$
But as $n_1$ is assumed to be a root, $(3n_1^3 -39n_1^2+360n_1 + 20) = 0$ and
$3n^3 - 39n^2 + 360n +20 = 3(n-n_1)(n^2+(n_1-13)n+(n_1^2-13n_1+120))$
And $\frac {3n^3 - 39n^2 + 360n +20}{3(n-n_1)} = n^2+(n_1-13)n+(n_1^2-13n_1+120)$
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