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factor completely. $y^2 - 6y + 5$

Question

factor completely.
$y^2 - 6y + 5$

Explanation:

Step1: Find two numbers

We need two numbers that multiply to \(5\) (the constant term) and add up to \(-6\) (the coefficient of the middle term). The numbers are \(-1\) and \(-5\) since \((-1)\times(-5) = 5\) and \((-1)+(-5)=-6\).

Step2: Factor the quadratic

Using the numbers from Step 1, we can factor \(y^{2}-6y + 5\) as \((y - 1)(y - 5)\).

Answer:

\((y - 1)(y - 5)\)