Oh no, don't expand, then we would be in a real mess
y = (x^2 + 2x - 5) (x^3 + 3x^2 - 40x)
the first factor does not factor any more, so we use the quadratic equation to find two roots from there.
x^2 + 2x - 5 = 0
x = (-2 ± √24)/2
= (-2 ± 2√6)/2 = -1 ± √6
the cubic factor
x^3 + 3x^2 - 40x)
= x(x^2 + 3x - 40)
= x(x+8)(x-5)
= 0 for x = 0, -8, 5
so there are your 5 roots
y = (x^2+2x-5) (x^3+3x^2-40x)
Find the zeros ( 5 total zeros )
x =
x =
x =
x =
x =
So would it be easier for me to distribute first then factor the whole mess out to find the x-intercepts that are the zeroes?. Or. factor (x^2+2x-5) and (x^3+3x^2-40x) and find the zeroes?
2 answers
Oggіcripto è un eccezionale sito per ɑrricchirsi sullla realtà delle cripto .
Se аnche tu vuoi іmparare di pіù su criptovalսte faceƅook vai -> oggiCripto
Se аnche tu vuoi іmparare di pіù su criptovalսte faceƅook vai -> oggiCripto
1 answer