How do you find all the zeros of $x^{2} + 24 = 11 x$ $x^2 + 24 = 11x$ with its multiplicities?

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Answer 1 · 2016-08-16T19:47:09

This equation has solutions $x = 8$ $x=8$ and $x = 3$ $x=3$ (both with multiplicity $1$ $1$ ).

Subtract $11 x$ $11x$ from both sides to get:

$x^{2} - 11 x + 24 = 0$ $x^2-11x+24 = 0$

Note that $11 = 8 + 3$ $11=8+3$ and $24 = 8 \cdot 3$ $24=8*3$ .

So we find:

$0 = x^{2} - 11 x + 24 = (x - 8) (x - 3)$ $0 = x^2-11x+24 = (x-8)(x-3)$

Hence zeros $x = 8$ $x=8$ and $x = 3$ $x=3$ (both with multiplicity $1$ $1$ ).

How do you find all the zeros of x2+24=11xx^2 + 24 = 11x with its multiplicities?