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Solve the equation \(2 x^{3}-5 x^{2}+x+2=0\) given that 2 is a zero of \(f(x)=2 x^{3}-5 x^{2}+x+2\)

Short Answer

Expert verified
The solution to the equation \(2 x^{3}-5 x^{2}+x+2=0\) are \(x=2, 2, \frac{1}{4}\).

Step by step solution

01

Confirm the Given Zero

The first step is to confirm that the given number is a zero of the function. This is done by substituting the given number into the equation to check if it equals zero. So we substitute \(x=2\) into the equation \(f(x)=2x^{3}-5x^{2}+x+2\) and if \(f(2)=0\), then indeed 2 is a zero of the equation. Checking this in detail leads to \(f(2)=2(2^{3})-5(2^{2})+(2)+2 = 0\). Therefore, 2 is a zero of the equation.
02

Perform Synthetic Division

Perform synthetic division, which is a shorthand version of polynomial division, using the given zero to reduce the equation to a quadratic. The process of synthetic division is performed as follows: List the coefficients of the equation (2, -5, 1, and 2), then write down the given zero. Bring the first coefficient directly down and multiply it by the given zero, then add this to the next coefficient. Repeat these operations until the end of the coefficients' list. The final number should be zero if the given zero is indeed a root. Doing this, we get \(2x^{2}-9x+4=0\).
03

Solve the Quadratic Equation

We obtain a quadratic equation from the synthetic division which we can solve It using the quadratic formula \(x = \frac{-b ± \sqrt{b^{2}-4ac}}{2a}\), where \(a\), \(b\) and \(c\) are coefficients of the equation. In our quadratic equation \(a=2\), \(b=-9\), \(c=4\). Plugging these values into the formula, we obtain \(x=\frac{9\pm\sqrt{81-32}}{4}\). Simplifying, this gives \(x=2, \frac{1}{4}\). Thus, the three solutions of the cubic equation are \(2, 2, \frac{1}{4}\).

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