SOLUTION: Find the roots of the polynomial function.
y=x^3+8
y=x^3+8=(x+2)(x^2-2x+4)
(-2)^2-4(1)(4)=12
one real root= -2
Could someone please check this for me?
Algebra ->
Radicals
-> SOLUTION: Find the roots of the polynomial function.
y=x^3+8
y=x^3+8=(x+2)(x^2-2x+4)
(-2)^2-4(1)(4)=12
one real root= -2
Could someone please check this for me?
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Question 187610This question is from textbook saxon algebra 2
: Find the roots of the polynomial function.
y=x^3+8
y=x^3+8=(x+2)(x^2-2x+4)
(-2)^2-4(1)(4)=12
one real root= -2
Could someone please check this for me? This question is from textbook saxon algebra 2
You can put this solution on YOUR website! Find the roots of the polynomial function.
y=x^3+8
y=x^3+8=(x+2)(x^2-2x+4)
(-2)^2-4(1)(4)=12
one real root= -2
Could someone please check this for me?
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y = (-2)^3 + 8
y = -8 + 8 = 0
It fits.
You don't need the complex roots? It says "Find the roots..."
You can put this solution on YOUR website! Find the roots of the polynomial function.
y=x^3+8
y=x^3+8=(x+2)(x^2-2x+4)
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x+2=0 or x^2-2x+4 = 0
x = -2 or Use the quadratic Formula:
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x = [2 +- sqrt(4-4*1*4)]/2
x = [2 +- sqrt(-12)]/2
x = [2 +- 2isqrt(3)]/2
x = [1 +- isqrt(3)
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Cheers,
Stan H.