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c(x)=2x^3+3x^2-1 find rational zeros

Sagot :

ivycoveredwalls

Answer:

1/2

Step-by-step explanation:

The Rational Root Theorem says that any rational root has

1. a numerator that is a factor of the constant term

2. a denominator that is a factor of the leading coefficient (that's attached to the highest-power term)

Numerator:  [tex]\pm 1[/tex]

Denominator: [tex]\pm 2[/tex]

That means there are only two possible rational roots: [tex]\pm \frac{1}{2}[/tex]

Try them both by plugging them into the polynomial.

[tex]2\left(\frac{1}{2}\right)^3 + 3\left(\frac{1}{2}\right)^2 -1= \frac{1}{4}+\frac{3}{4}-1 =0[/tex]

Aha!  The negative one-half value does not produce 0

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