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Question
question 25 of 25
factor this trinomial.
$x^2 - 8x + 15$
a. $(x - 3)(x - 5)$
b. $(x - 3)(x + 5)$
c. $(x - 1)(x + 15)$
d. $(x - 1)(x - 15)$
Step1: Recall factoring trinomial form \(x^2 + bx + c\)
We need two numbers that multiply to \(15\) (the constant term) and add up to \(-8\) (the coefficient of \(x\)).
Step2: Find the two numbers
The factors of \(15\) are \(1\) & \(15\), \(3\) & \(5\). Since the sum is \(-8\) and product is \(15\), the numbers are \(-3\) and \(-5\) (because \((-3)\times(-5)=15\) and \((-3)+(-5)=-8\)).
Step3: Factor the trinomial
Using the numbers \(-3\) and \(-5\), we can factor \(x^2 - 8x + 15\) as \((x - 3)(x - 5)\). We can check by expanding \((x - 3)(x - 5)=x^2-5x-3x + 15=x^2-8x + 15\), which matches the original trinomial.
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A. \((x - 3)(x - 5)\)