# How do you write a quadratic equation in standard form with solutions 1 and -8?

Oct 7, 2016

The roots of a polynomial are, in a certain sense, the blocks to build the polynomial itself.

Take for example an expression like $f \left(x\right) = \left(x - \setminus \alpha\right)$, for some real number $\setminus \alpha$. This is a line (first degree polynomial) which equals zero exacly when $x = \setminus \alpha$, since $f \left(\setminus \alpha\right) = \setminus \alpha - \setminus \alpha = 0$.

To build a polynomial with higher degree, simply multiply such blocks: if we consider $g \left(x\right) = \left(x - \setminus \alpha\right) \left(x - \setminus \beta\right)$, this function will have exactly $\setminus \alpha$ and $\setminus \beta$ as roots, since now we have

$g \left(\setminus \alpha\right) = \left(\setminus \alpha - \setminus \alpha\right) \left(\setminus \alpha - \setminus \beta\right) = 0 \cdot \left(\setminus \alpha - \setminus \beta\right) = 0$
$g \left(\setminus \beta\right) = \left(\setminus \beta - \setminus \alpha\right) \left(\beta - \setminus \beta\right) = \left(\setminus \beta - \setminus \alpha\right) \cdot 0 = 0$

In this way, you can build a polynomial of the degree you want, with the roots that you want. In your case, you have to build

$p \left(x\right) = \left(x - 1\right) \left(x + 8\right)$

In fact,

$p \left(1\right) = \left(1 - 1\right) \left(1 + 8\right) = 0 \cdot 9 = 0$
$p \left(- 8\right) = \left(- 8 - 1\right) \left(- 8 + 8\right) = - 9 \cdot 0 = 0$

So, $p \left(x\right)$ has the roots we want it to have. To write it in standard form, simply expand the parenthesis:

$\left(x - 1\right) \left(x + 8\right) = {x}^{2} + 7 x - 8$

Faster Strategy:

In the particular case of quadratic polynomials, we have a faster way to solve exercises like this: the generic expression

${x}^{2} + a x + b$ (note that the ${x}^{2}$ coefficient must be $1$)

yields two solutions ${x}_{1}$ and ${x}_{2}$ such that

${x}_{1} + {x}_{2} = - a$
${x}_{1} {x}_{2} = b$

In other words, the sum of the solutions is the opposite of the $x$ coefficient, and the product of the solutions is the last coefficient of the quadratic expression.

If your solutions have to be $1$ and $- 8$, then we have

$a = - \left(1 - 8\right) = 7$
$b = 1 \cdot \left(- 8\right) = - 8$

and again we come to the conclusion that

$p \left(x\right) = {x}^{2} + 7 x - 8$