Find an ordered pair (a,b) such that the polynomial f(x)=x^3+(a - 1)x^2+(b+3)x+1 is divisible by x^2 - 1.

Guest Sep 30, 2020

#1**0 **

The top row is what's being multiplied to the left row.

Let's fill in information that we already know. (I left out the values that have variables in the coefficients because we don't know their values yet).

\(x^2\) | \(-1\) | |

\(x^3\) | ||

\(1\) |

We know that dividing \(\frac{x^3}{x^2} = x\) is true, so let's fill this in our table.

\(x^2\) | \(-1\) | |

\(x\) | \(x^3\) | |

\(1\) |

Now, we know that \(\frac{1}{-1} = -1\) is true, so let's fill this in our table as well.

\(x^2\) | \(-1\) | |

\(x\) | \(x^3\) | |

\(-1\) | \(1\) |

Using what we have, we can fill in the rest of the table.

\(x^2\) | \(-1\) | |

\(x\) | \(x^3\) | \(-1x\) |

\(-1\) | \(-1x^2\) | \(1\) |

With our new values, we can find \((a, b)\). We know that \((a-1)x^2 = -1x^2\) is true, so solving for \(a\), we get \(a = 0\). Similarly we can say that \((b+3)x = -1x\) is true, so solving for \(b\), we get \(b = -4\).

Our final answer is \((a,b) = \boxed{(0,-4)}\).

Guest Sep 30, 2020