Factors of the Polynomial of Degree 4
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Why does the polynomial of degree 4 have the following factors?
$$a+b-c $$ $$a-b+c $$ $$-a+b+c $$ $$a+b+c $$ -
@aaron-wang said in Factors of the Polynomial of Degree 4:
a+b+c
Oh, wow, this is a nice question! (By the way, only the first three rows are factors of the \(\text{area}^2\) formula.... not the \(a + b + c.\) )
The fourth-degree polynomial that Prof. Loh is speaking about is the \(\text{(area)}^2\) formula. He writes an ugly-looking expression down. But he's not looking at this ugly formula when he realizes that \( a + b - c, a - b + c,\) and \(-a + b + c\) are factors. He's just thinking about what the area of a triangle would be like if two of the sides added together were to equal the third side. (No algebra involved!!
)
You see, if \(a + b - c = 0,\) then you can't make a triangle. You just get two straight lines that line up along the third line. And we call a "root" of a polynomial another expression that makes the polynomial \(0.\)
Like for example, consider \(x^2 + 5x + 6 = 0.\) When \(x = 3,\) then \(x^2 + 5x + 6 = 0,\) so \(\textcolor{red}{x -3}\) is a root of that polynomial. You can even factor the polynomial out as \( (\textcolor{red}{x-3})(x-2) = 0.\)
So we know that the area formula looks something like:
\( \text{ Area}^2 = (a + b -c)(a - b + c)(-a + b + c) \times \text{(some other stuff....)}.\)
Because when one of those terms (factors) equals \(0,\) then the area equals \(0.\)
I hope this helped. Please let me know if there's something I can clarify for you! Thanks again for asking a great question!
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@debbie Ah, I see! Thank you for clarifying this, it definitely helped.
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@aaron-wang That's great!
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