What are all the possible rational zeros for $f (x) = x^{2} + 8 x + 10$ $f(x)=x^2+8x+10$ ?

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What are all the possible rational zeros for $f (x) = x^{2} + 8 x + 10$ $f(x)=x^2+8x+10$ ?

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Answer 1 · 2018-03-28T16:13:49

The roots are $- 4 + \sqrt{6}$ $-4+\sqrt{6}$ and $- 4 - \sqrt{6}$ $-4-\sqrt{6}$ .

Explanation:

You start with $f (x) = x^{2} + 8 x + 10$ $f(x)=x^2+8x+10$ . First you have to establish its degree, which is $2$ $2$ because the highest exponent is $2$ $2$ . That being said, you can apply the quadratic formula, namely:

$a x^{2} + b x + c = 0 \Rightarrow x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$ $ax^2+bx+c=0\implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

In this particular case, $a$ $a$ is $1$ $1$ , $b$ $b$ is $8$ $8$ and $c$ $c$ is $10$ $10$ , and therefore:

$x = \frac{- 8 \pm \sqrt{8^{2} - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$ $x=\frac{-8\pm\sqrt{8^2-4\cdot 1\cdot 10}}{2\cdot 1}$

$x = \frac{- 8 \pm \sqrt{64 - 40}}{2}$ $x=\frac{-8\pm\sqrt{64-40}}{2}$

$x = \frac{- 8 \pm \sqrt{24}}{2} = \frac{- 8 \pm \sqrt{2 \cdot 2 \cdot 6}}{2}$ $x=\frac{-8\pm\sqrt{24}}{2}=\frac{-8\pm\sqrt{2\cdot 2\cdot 6}}{2}$

$x = \frac{- 8 \pm 2 \sqrt{6}}{2} = - 4 \pm \sqrt{6}$ $x=\frac{-8\pm2\sqrt{6}}{2}=-4\pm\sqrt{6}$

Hence, the two roots are $- 4 + \sqrt{6}$ $-4+\sqrt{6}$ and $- 4 - \sqrt{6}$ $-4-\sqrt{6}$ .

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What are all the possible rational zeros for $f (x) = x^{2} + 8 x + 10$ $f(x)=x^2+8x+10$ ?

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