SOLUTION: how do you find all of the real zeroes of 8x^3+125

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Question 709135: how do you find all of the real zeroes of 8x^3+125
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
8x%5E3%2B125
To find the zeroes of an expression of degree 3 or more (like this one) we need to factor it.

When factoring, check the greatest common factor (GCF) first. The GCF here is 1 (which we rarely factor out).

After the GCF, we look to use any and all of the other factoring techniques. One of these techniques is factoring by the use of factoring patterns. And one of these patterns (the only one in fact) that has two terms with a "+" between them is:
a%5E3%2Bb%5E3+=+%28a%2Bb%29%28a%5E2-ab%2Bb%5E2%29
As the pattern shows, we need a sum of two cubes to use this pattern. Is 8x%5E3%2B125 a sum of cubes? It's definitely a sum. Are 8x%5E3 and 125 perfect cubes? With a little investigation we should find that 8x%5E3+=+%282x%29%5E3 and 125+=+5%5E3. So we can use the pattern with the "a" being "2x" and the "b" being "5". So 8x%5E3%2B125, according to the patterns factors into:
%28%282x%29%2B%285%29%29%28%282x%29%5E2-%282x%29%285%29%2B%285%29%5E2%29
(Note: My use of parentheses may seem excessive. But it is actually a good idea to use parentheses like this when making substitutions like we have done here. It helps us avoid easily-made mistakes.)
This simplifies to:
%282x%2B5%29%284x%5E2-10x%2B25%29
Neither factor will factor further no matter which techniques we try.

From the Zero Product Property we know that this product will be zero only be zero if one of the factors is zero. And if this product is zero then the expression from which we got these factors, 8x%5E3%2B125 will also be zero. So the x values that make these factors zero will be the zeroes of 8x%5E3%2B125. Setting each factor to zero and then solving will lead to our solution. For the first factor:
2x%2B5+=+0
Solving this we should get x = -5/2.

For the second factor:
4x%5E2-10x%2B25+=+0
We must use the Quadratic Formula:
x+=+%28-%28-10%29+%2B-+sqrt%28%28-10%29%5E2-4%284%29%2825%29%29%29%2F2%284%29
simplifying...
x+=+%28-%28-10%29+%2B-+sqrt%28100-4%284%29%2825%29%29%29%2F2%284%29
x+=+%28-%28-10%29+%2B-+sqrt%28100-400%29%29%2F2%284%29
x+=+%28-%28-10%29+%2B-+sqrt%28-300%29%29%2F2%284%29
At this point we can stop. The negative in the square root means that the solutions will be complex numbers. But the problem specifically asks for real solutions. So the only real zero to 8x%5E3%2B125 is -5/2.

P.S. In case you are curious about the complex zeroes I will go ahead and finish with the Quadratic Formula. Continuing the simplifying:
x+=+%2810+%2B-+sqrt%28-300%29%29%2F8
x+=+%2810+%2B-+sqrt%28-1%2A100%2A3%29%29%2F8
x+=+%2810+%2B-+sqrt%28-1%29%2Asqrt%28100%29%2Asqrt%283%29%29%2F8
x+=+%2810+%2B-+i%2A10%2Asqrt%283%29%29%2F8
x+=+%282%285+%2B-+5i%2Asqrt%283%29%29%29%2F%282%2A4%29
x+=+%285+%2B-+5i%2Asqrt%283%29%29%2F4
which is short for:
x+=+%285+%2B+5i%2Asqrt%283%29%29%2F4 or x+=+%285+-+5i%2Asqrt%283%29%29%2F4
In standard "a + bi" form these would be:
x+=+5%2F4+%2B+%285sqrt%283%29%2F4%29i or x+=+5%2F4+%2B+%28-5sqrt%283%29%2F4%29i
So there are three zeroes: two complex (above) and the one real we found earlier (-5/2).