6 July, 23:18
Factor completely: x^4-14x^2+32
7 July, 00:03
x⁴ - 14x² + 32
x = - (-14) ± √ (-14) ² - 4 (1) (32) / 2 (1)
x = 14 ± √196 - 128 / 2
x = 14 ± √68 / 2
x = 14 ± 2√17 / 2
Factor the numerator
x = 2 (7 ± √17) / 2
Cancel out the two on both the numerator and denominator.
x = 7 ± √17
The two zeros for this problem are 7 ± √17.
So far, the zeros are irrational. These roots are real, unequal, irrational roots. The zeros are not perfect squares.
If you graph this polynomial function into a graphing calculator, you'll see that the parabola doesn't intersect the x-axis at rational values.
Furthermore, I have tried to factor this polynomial function by doing the AC method and grouping. However, the closest factorization I got was:
(x² - 12) (x² - 2.667) which is not even equivalent to the given function.
Although this function has real zeros, I believe that it is impossible to factor. Therefore, my answer is that this function is not factorable.
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» Factor completely: x^4-14x^2+32