Find the roots of the polynomial \(q(\lambda)\) and their algebraic multiplicities if
| \(q(\lambda)\) | \(=\) | \(\lambda^3 +4\lambda^2-3\lambda-18\) |
First we need to find the roots of \(q\), that is those number values for \(\lambda\) with \(q(\lambda)=0\).
By trial and error we find
| \(q(2)\) | \(=\) | \(2^3 + 4\cdot 2^2 - 3\cdot 2-18\) |
| \(\) | \(=\) | \(0\) |
So \(\lambda_1\DefEq 2\) is a root of \(q\). So we know that \((\lambda-2)\) divides \(q\). Via long division we find
| \(q(\lambda)\) | \(=\) | \((\lambda-2)(\lambda^2 +6\lambda +9)\) |
| \(\) | \(=\) | \((\lambda-2)^1(\lambda+3)^2\) |
We conclude that \(q\) has
Find the algebraic multiplicities of the roots of the polynomial
| \(p(\lambda)\) | \(=\) | \((\lambda-3)^5(\lambda+1)^2\) |
Here we get lucky: \(p\) is given to us in a form from which we can read off its roots and their multiplicities right away.
Indeed, the product \((\lambda-3)^5(\lambda+1)^2\) is \(0\) exactly when at least one of its factors \((\lambda-3)\) or \((\lambda+1)\) is \(0\).
Now, \(\lambda-3=0\) exactly when \(\lambda=3\). So \(p\) has the root \(\lambda_1 = 3\) with algebraic multiplicity \(5\).
Similarly, \(\lambda+1=0\) exactly when \(\lambda=-1\). So \(p\) has the root \(\lambda_2=-1\) with algebraic multiplicity \(2\).