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Find the complex zeros of each polynomial function f(x). Write f in factored form.

f(x)=2x4+2x3-11x2+x-6

Short Answer

Expert verified

The factored form isf(x)=(x+3)(x-2)(x+22i)(x-22i)

Step by step solution

01

Step 1. Given Information

The given function isf(x)=2x4+2x3-11x2+x-6

02

Step 2. Explanation 

The given function has a degree 4. So, there will be a maximum of 4 complex zeroes.

We will list the integers which are factors of the constant term -6.

p=±1,±2,±3,±6

Now, we will list the integers which are factors of the leading coefficient 2.

q=±1,±2

Next, list the possible potential rational zeroes which is the ratio pq

pq=±62,±32,±22,±12,±61,±31,±21,±11=±6,±2,±3,±1,±32,±12

03

Step 3. Graph

Graph the function.

From the above graph, the function intersects x-axis at two distinct points which are x=-3andx=2

Thus, the two factors are(x+3),(x-2)

04

Step 4. Calculation   

Use the synthetic division and divide the polynomial by (x+3)

Thus, the depressed equation is 2x3-4x2+x-2=0

Now, factorize the above equation by grouping.

2x3-4x2+x-2=02x2(x-2)+1(x-2)=0(2x2+1)(x-2)=0x-2=0⇒x=22x2+1⇒x=-12⇒x=±22i

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