The original question is find all the x intercepts (if any) for the following polynomial:
f(x) = (x^8 + 5) (x^12 + 7)
When I factored this polynomial I got: x^20 + 7x^8 + 5x^12 +35
How do I convert this into the correct format: ax^2 + bx + c to find the x intercepts? Or would the answer be, "No x intercepts" because you can not convert that polynomial into the correct format?
Please help me solve the original question. Thanks!
f(x) = (x^8 + 5) (x^12 + 7)
When I factored this polynomial I got: x^20 + 7x^8 + 5x^12 +35
How do I convert this into the correct format: ax^2 + bx + c to find the x intercepts? Or would the answer be, "No x intercepts" because you can not convert that polynomial into the correct format?
Please help me solve the original question. Thanks!
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That ain't anywhere near a quadratic!
They actually were quite nice to you in factoring it for you. Look at each factor.
x^8 + 5 is ALWAYS positive because x^8 is always 0 or greater.
x^12 + 7 is ALWAYS positive because x^12 is always 0 or greater.
Therefore, f(x) is always the product of two positve numbers. Therefore, there ARE no real zeros or x-intercepts of this polynomial!
Important points:
1) x^2, x^4, x^6, etc. are ALWAYS 0 or greater. Even exponent <===> can't be negative!
2) Sum of a nonnegative and a positive number will always be positive. (I glossed over this in my explanation, but since x^8 is 0 or greater, x^8 + 5 is 5 or greater, which means it's greater than zero.)
3) Product of two positives is always positive.
It has nothing to do with jamming the equation into the form of a quadratic...although that can be VERY helpful for equations like x^4 + x^2 + 9 = 0. (Can you see quickly, however, that this equation has no real solutions? Hint: once again, look at the signs of x^4 and x^2 and 9!)
Another way to look at this problem is that they gave it to you in the form of ab = 0. Thus, a = 0 or b = 0. In this case, we have x^8 + 5 = 0 or x^12 + 7 = 0. Rearranging the first gives x^8 = -5, which is impossible...anything to the 8th is positive. Rearranging the second gives x^12 = -7, and you get the same argument.
Important point: this only works for EVEN exponents! That's because if x is negative, x^8 is multiplying 8 negative numbers together. If you have the expression x^9, it can be either positive or negative...in fact, it's the same sign as x.
They actually were quite nice to you in factoring it for you. Look at each factor.
x^8 + 5 is ALWAYS positive because x^8 is always 0 or greater.
x^12 + 7 is ALWAYS positive because x^12 is always 0 or greater.
Therefore, f(x) is always the product of two positve numbers. Therefore, there ARE no real zeros or x-intercepts of this polynomial!
Important points:
1) x^2, x^4, x^6, etc. are ALWAYS 0 or greater. Even exponent <===> can't be negative!
2) Sum of a nonnegative and a positive number will always be positive. (I glossed over this in my explanation, but since x^8 is 0 or greater, x^8 + 5 is 5 or greater, which means it's greater than zero.)
3) Product of two positives is always positive.
It has nothing to do with jamming the equation into the form of a quadratic...although that can be VERY helpful for equations like x^4 + x^2 + 9 = 0. (Can you see quickly, however, that this equation has no real solutions? Hint: once again, look at the signs of x^4 and x^2 and 9!)
Another way to look at this problem is that they gave it to you in the form of ab = 0. Thus, a = 0 or b = 0. In this case, we have x^8 + 5 = 0 or x^12 + 7 = 0. Rearranging the first gives x^8 = -5, which is impossible...anything to the 8th is positive. Rearranging the second gives x^12 = -7, and you get the same argument.
Important point: this only works for EVEN exponents! That's because if x is negative, x^8 is multiplying 8 negative numbers together. If you have the expression x^9, it can be either positive or negative...in fact, it's the same sign as x.