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# Factor trinomial $$ \color{blue}{ x^2-3x-40 } $$

## Answer

The factored form is $$ \color{blue}{ x^2-3x-40 = \left(x+5\right)\left(x-8\right) } $$

## Explanation

** Step 1:** Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$.
( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -3 } ~ \text{ and } ~ \color{red}{ c = -40 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -3 } $ and multiply to $ \color{red}{ -40 } $.

** Step 2:** Find out pairs of numbers with a product of $\color{red}{ c = -40 }$.

PRODUCT = -40 |

-1 40 | 1 -40 |

-2 20 | 2 -20 |

-4 10 | 4 -10 |

-5 8 | 5 -8 |

** Step 3:** Find out which pair sums up to $\color{blue}{ b = -3 }$

PRODUCT = -40 and SUM = -3 |

-1 40 | 1 -40 |

-2 20 | 2 -20 |

-4 10 | 4 -10 |

-5 8 | 5 -8 |

** Step 4:** Put 5 and -8 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-3x-40 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-3x-40 & = (x + 5)(x -8) \end{aligned} $$