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Hi!! I found me in trouble when I saw the solution of a simple inequality, that can be found at the end of the first chapter, that is the exercise 4 - (viii): $x^2+x+1 > 0$. Very easy to solve I know, but I also saw the related solution on the Spivak Calculus Answer Book and I did not understand how succeed to factoring the polynomial ( already factored into the solution ), I show what I mean:

$$x^2 + x + 1 > 0 = \left( x + \frac{ 1}{2} \right)^2 + \frac{3}{4}$$

On the RHS there is the factored expression and now my question is: how can I get this form for every second degree polynomial? Is there a specific formula to get this form?

I hope someone could help me

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    $\begingroup$ Look up: completing the square. $\endgroup$ – Zain Patel Jul 20 '15 at 19:06
  • $\begingroup$ Among the various methods for motivating how one might discover how to complete the square, here's one you might not come across. Note that $x^2+x+1=x(x+1)+1,$ where the two variable factors are "not balanced" (the zeros of $x$ and $x+1$ are $0$ and $-1,$ which are not symmetric with respect to the origin). To balance the variables, change variables by letting $x=u-\frac{1}{2}$ and $x+1=u+\frac{1}{2},$ so that $x(x+1)+1$ becomes $(u-\frac{1}{2})(u+\frac{1}{2}) + 1 = u^2-\frac{1}{4}+1=u^2+\frac{3}{4}.$ Now convert back to $x$'s using $u=x+\frac{1}{2}.$ $\endgroup$ – Dave L. Renfro Jul 20 '15 at 19:56
  • $\begingroup$ By the way, the expression you got is not "factored". Factoring an expression generally means to reduce it to a product of polynomial factors each of lesser degree than the original expression. The expression $x^2 + x + 1$ cannot be factored in the reals, however, it is possible to find a factorisation in the complex numbers, i.e. $x^2 + x + 1 = (x - \omega)(x - \overline{\omega})$, where $\omega$ and its conjugate are complex cube roots of $1$. Completing the square is not the same as factorisation, however, it can help you simplify expressions and solve equations/inequalities. $\endgroup$ – Deepak Jul 21 '15 at 1:37
  • $\begingroup$ Thank you all for the answers and the comments! Now everything's clear! $\endgroup$ – Michele Jul 21 '15 at 21:24
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Completing the square is the process you're looking for. Basically, for a quadratic $ax^2 + bx + c = a(x^2 + \frac{b}{a}x + \frac{c}{a})$. The term inside the bracket can be written as $$\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a}$$ so that for any general quadratic expression we have $$ax^2 + bx + c\equiv a\left(\left(x + \frac{b}{2a}\right)^2 - \left(\frac{b}{2a}\right)^2 + \frac{c}{a}\right ).$$

In fact, writing it in this form is a useful tool because $x = -\frac{b}{2a}$ is the minimum/maximum point of the quadratic.

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As suggested, this was achieved through completing the square:

$$ x^2 +x+1 = x^2 + x +\left(\frac{1}{2}\right)^2 + 1 -\frac{1}{4}$$

$$ = \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}$$

You need to go and read up completing the square to understand how I got the above.

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$$ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(\left(x+\frac{b}{2a}\right)^2+\frac{c}{a}-\frac{b^2}{4a^2}\right)\\=a\left(x+\frac{b}{2a}\right)^2+\frac{b^2-4ac}{a}$$

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  • $\begingroup$ (I think maybe you need to adjust the last term in your answer.) $\endgroup$ – user84413 Jul 20 '15 at 22:20
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For $a\not =0$, we have$$\begin{align}ax^2+bx+c&=a\left(x^2+\frac{b}{a}x\right)+c\\&=a\left(\color{red}{x^2+\frac{b}{a}x+\left(\frac{b}{2a}\right)^2}-\left(\frac{b}{2a}\right)^2\right)+c\\&=a\left(\color{red}{\left(x+\frac{b}{2a}\right)^2 }-\left(\frac{b}{2a}\right)^2\right)+c\\&=a\left(x+\frac{b}{2a}\right)^2-a\cdot\frac{b^2}{4a^2}+c\\&=a\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a}+c\end{align}$$

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We see that $x^2+x+1=(x^2+x+{1\over 4})+{3\over 4}=(x+{1\over 2})^2+{3\over4}$. Since $(x+{1\over 2})^2\ge 0$ for all $x\in \mathbb{R}$, then it follows that $(x+{1\over 2})^2+{3\over4}>0$ and so $x^2+x+1>0$.

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Simply you have to add and subtract the square of half of $x$'s coefficient then the algebraic statement will be written as sum or difference of to square.

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