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Question
complex numbers online practice
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what are the roots of the polynomial function $p(x) = x^3 + 4x^2 - 15x - 18$? (1 point)
\\(\boldsymbol{x = -3, x = -1}\\) and \\(\boldsymbol{x = 6}\\)
\\(\boldsymbol{x = -6, x = 1}\\) and \\(\boldsymbol{x = 3}\\)
\\(\boldsymbol{x = 3, x = 1}\\) and \\(\boldsymbol{x = 6}\\)
\\(\boldsymbol{x = -6, x = -1}\\) and \\(\boldsymbol{x = 3}\\)
Step1: Test possible rational roots
By Rational Root Theorem, possible roots are factors of 18 over factors of 1, so ±1, ±2, ±3, ±6, ±9, ±18. Test \( x = -1 \): \( p(-1)=(-1)^3 + 4(-1)^2 - 15(-1) - 18=-1 + 4 + 15 - 18 = 0 \). So \( x = -1 \) is a root.
Step2: Factor the polynomial
Since \( x = -1 \) is a root, \( (x + 1) \) is a factor. Divide \( p(x) \) by \( (x + 1) \) using polynomial division or synthetic division. Using synthetic division for \( x=-1 \):
So \( p(x)=(x + 1)(x^2 + 3x - 18) \).
Step3: Factor the quadratic
Factor \( x^2 + 3x - 18 \): find two numbers that multiply to -18 and add to 3. They are 6 and -3. So \( x^2 + 3x - 18=(x + 6)(x - 3) \).
Step4: Find all roots
Thus, \( p(x)=(x + 1)(x + 6)(x - 3) \). Setting each factor to zero: \( x + 1 = 0 \Rightarrow x = -1 \), \( x + 6 = 0 \Rightarrow x = -6 \), \( x - 3 = 0 \Rightarrow x = 3 \).
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\( x = -6 \), \( x = -1 \) and \( x = 3 \) (corresponding to the option: \( x = -6, x = -1 \) and \( x = 3 \))