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| − | '''Vieta's Formulas''', otherwise called Viète's Laws, are a set of [[equation]]s relating the [[root]]s and the [[coefficient]]s of [[polynomial]]s.
| + | #REDIRECT[[Vieta's formulas]] |
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| − | == Introduction ==
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| − | Vieta's Formulas were discovered by the French mathematician [[François Viète]].
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| − | Vieta's Formulas can be used to relate the sum and product of the roots of a polynomial to its coefficients. The simplest application of this is with quadratics. If we have a quadratic <math>x^2+ax+b=0</math> with solutions <math>p</math> and <math>q</math>, then we know that we can factor it as
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| − | <center><math>x^2+ax+b=(x-p)(x-q)</math></center>
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| − | (Note that the first term is <math>x^2</math>, not <math>ax^2</math>.) Using the distributive property to expand the right side we get
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| − | <center><math>x^2+ax+b=x^2-(p+q)x+pq</math></center>
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| − | We know that two polynomials are equal if and only if their coefficients are equal, so <math>x^2+ax+b=x^2-(p+q)x+pq</math> means that <math>a=-(p+q)</math> and <math>b=pq</math>. In other words, the product of the roots is equal to the constant term, and the sum of the roots is the opposite of the coefficient of the <math>x</math> term.
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| − | A similar set of relations for cubics can be found by expanding <math>x^3+ax^2+bx+c=(x-p)(x-q)(x-r)</math>.
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| − | We can state Vieta's formula's more rigorously and generally. Let <math>P(x)</math> be a polynomial of degree <math>n</math>, so <math>P(x)={a_n}x^n+{a_{n-1}}x^{n-1}+\cdots+{a_1}x+a_0</math>,
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| − | where the coefficient of <math>x^{i}</math> is <math>{a}_i</math> and <math>a_n \neq 0</math>. As a consequence of the [[Fundamental Theorem of Algebra]], we can also write <math>P(x)=a_n(x-r_1)(x-r_2)\cdots(x-r_n)</math>, where <math>{r}_i</math> are the roots of <math>P(x)</math>. We thus have that
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| − | <center><math> a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0 = a_n(x-r_1)(x-r_2)\cdots(x-r_n).</math></center>
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| − | Expanding out the right hand side gives us
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| − | <math> a_nx^n - a_n(r_1+r_2+\!\cdots\!+r_n)x^{n-1} + a_n(r_1r_2 + r_1r_3 +\! \cdots\! + r_{n-1}r_n)x^{n-2} +\! \cdots\! + (-1)^na_n r_1r_2\cdots r_n.</math>
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| − | The coefficient of <math> x^k </math> in this expression will be the <math>k </math>th [[symmetric sum]] of the <math>r_i</math>.
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| − | We now have two different expressions for <math>P(x)</math>. These must be equal. However, the only way for two polynomials to be equal for all values of <math>x</math> is for each of their corresponding coefficients to be equal. So, starting with the coefficient of <math> x^n </math>, we see that
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| − | <center><math>a_n = a_n</math></center>
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| − | <center><math> a_{n-1} = -a_n(r_1+r_2+\cdots+r_n)</math></center>
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| − | <center><math> a_{n-2} = a_n(r_1r_2+r_1r_3+\cdots+r_{n-1}r_n)</math></center>
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| − | <center><math>\vdots</math></center>
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| − | <center><math>a_0 = (-1)^n a_n r_1r_2\cdots r_n</math></center>
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| − | More commonly, these are written with the roots on one side and the <math>a_i</math> on the other (this can be arrived at by dividing both sides of all the equations by <math>a_n</math>).
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| − | If we denote <math>\sigma_k</math> as the <math>k</math>th symmetric sum, then we can write those formulas more compactly as <math>\sigma_k = (-1)^k\cdot \frac{a_{n-k}}{a_n{}}</math>, for <math>1\le k\le {n}</math>.
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| − | ==Problems==
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| − | ===Beginner===
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| − | *Let <math>r_1,r_2,</math> and <math>r_3</math> be the three roots of the cubic <math>x^3+3x^2+4x-4</math>. Find the value of <math>r_1r_2+r_1r_3+r_2r_3</math>.
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| − | *Suppose the polynomial <math>5x^3+4x^2-8x+6</math> has three real roots <math>a,b</math>, and <math>c</math>. Find the value of <math>a(1+b+c)+b(1+a+c)+c(1+a+b)</math>.
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| − | ===Intermediate===
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| − | ===Olympiad===
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| − | == See Also ==
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| − | * [[Algebra]]
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| − | * [[Polynomials]]
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| − | * [[Newton's Sums]]
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| − | == External Links ==
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| − | *[http://mathworld.wolfram.com/VietasFormulas.html Mathworld's Article]
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| − | [[Category:Elementary algebra]]
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