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pantera1 [17]
3 years ago
8

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!! I CANNOT RETAKE THIS!!

Mathematics
1 answer:
irina1246 [14]3 years ago
8 0

5x^3 - 4x + 1 = 0 \\\\5x^3-5x+x+1=0\\\\5x(x^2-1)+1(x+1)=0\\\\5x(x+1)(x-1)+1(x+1)=0\\\\(x+1)(5x(x-1)+1)=0\\\\(x+1)(5x^2-5x+1)=0\\\\x=-1\\\\5x^2-5x+1=0\\\\\Delta=(-5)^2-4\cdot5\cdot1=25-20=5

\Delta>0 therefore there exist two real solutions.

In total, there are 3 real roots. But, real roots are also complex roots (doesn't work the other way round!), so there are 3 complex roots.

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Step-by-step explanation:

So using the commutative property, we can change the equation 15x^3-6x^2-25x+10 into 15x^3-25x -6x^2+10

Let’s split that into two sections so it’s easier to see:

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Next let’s look at what 15x^3 and -25x have in common. They have 5x in common.

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