Factoring Help (To Determine Zeros): f(x) = 2x^4 + x^2 - 15
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pka
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\(\displaystyle 2x^4 + x^2 - 15=(2x^2-5)(x^2+3)\)
HallsofIvy
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Notice that there are only even powers of x. If we let \(\displaystyle y= x^2\) then we can write the function as \(\displaystyle f(y)= 2y^2+ y- 15\). You could, if you had to, use the "quadratic formula" to determine the zeros of that. Or you could use the "rational root theorem"- any rational number zeros, a/b, must have numerator, a, that divides 15 and denominator, b, that divides 2. So any rational number zeros must be one of 1/2, -1/2, 1, -1, 3/2, -3/2, 5/2, -5/2, 3, -3, 15/2, -15/2, 15, or -15.
By evaluating the polynomial at those, we see that 5/2 and -3 are roots. We can factor \(\displaystyle 2y^2+ y- 15= (2y- 5)(y+ 3)\) and, replacing y with \(\displaystyle x^2\), \(\displaystyle 2x^4+ x^2- 15= (2x^2- 5)(x^2+ 3)\).
By evaluating the polynomial at those, we see that 5/2 and -3 are roots. We can factor \(\displaystyle 2y^2+ y- 15= (2y- 5)(y+ 3)\) and, replacing y with \(\displaystyle x^2\), \(\displaystyle 2x^4+ x^2- 15= (2x^2- 5)(x^2+ 3)\).