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Chapter 2: Problem 78

Write the polynomial as the product of linear factors and list all the zeros of the function. $$h(x)=x^{4}+6 x^{3}+10 x^{2}+6 x+9$$

Short Answer

Expert verified
The polynomial \(h(x) = x^{4} + 6x^{3} + 10x^{2} + 6x + 9\) when factorised as the product of linear factors becomes \((x^{2} + 3)^{2}\). The zeros of the function are complex numbers, specifically \(x = \pm i\sqrt{3}\)

Step by step solution

01

Factorising the polynomial

Here we'll factorise the polynomial function to its simplest form: \(h(x)=(x^{2}+3)^2\)
02

Setting the factorised polynomial equal to zero

Now, to find the zeros of the function, we set the factorised polynomial equal to zero: \((x^{2}+3)^2 = 0\
03

Solve for x

Next step involves solving the above equation for \(x\). Doing so will result in two potential solutions for \(x\), both of which are complex numbers: \(x = \pm i\sqrt{3}\)
04

Listing the Zeros of the Function

Finally, we list the zeros as the number(s) that make(s) the polynomial function equal to zero. From Step 3, we find these to be \(x = \pm i\sqrt{3}\)

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