Question

Write the polynomial as the product of linear factors.

h(x) = x^2 ? 2x + 10

Answer 1

If
`h(x) = x^2 + 2x + 10` then
`(x+1-3i)*(x+1+3i)`.

If
`h(x) = x^2 - 2x + 10` then
`(x-1-3i)*(x-1+3i)`.

Explanation:

In order to write the polynomial as the product of linear factors, we need to find the roots of the polynomial. A quadratic equation is defined as:

`ax^2+bx+c`

Because the given polynomial expression is a quadratic equation, we can use the following equations for calculating the roots:

`x1=(-b/2a)+√(b^2-4ac)/2a`

`x1=(-b/2a)-√(b^2-4ac)/2a`

Since the second term sign is not given, then we can write the expression as:

`x^2+s2x+10`, in which a=1, b=s2 where 's' represents a sign (- or +), and c=10.

Using the equation for finding the roots we obtain:

`x1=(-sb/2a)+√(b^2-4ac)/2a`

`x1=(-s2/2*1)+√(2^2-4*1*10)/(2*1)`; notice that
`(sb)^(2) = 2^(2)`

`x1=(-s1)+√(-36)/2`

`x1=-(s1)+6i/2`

`x1=-(s1)+3i`

`x2=(-sb/2a)-√(b^2-4ac)/2a`

`x2=(-s2/2*1)-√(2^2-4*1*10)/(2*1)`; notice that (s2)²=2^2

`x2=(-s1)-√(-36)/2`

`x2=-(s1)-6i/2`

`x2=-(s1)-3i`

If we consider 's' as possitive (+) the roots are:

`x1=-1+3i` and
`x2=-1-3i`

Whereas if we consider 's' as negative (-) the roots are:

`x1=1+3i` and
`x2=1-3i`

The above means that if the equation is
`h(x) = x^2 + 2x + 10`, then we can express the polynomial as:
`(x+1-3i)*(x+1+3i)`.

But, if the equation is
`h(x) = x^2 - 2x + 10`, then we can express the polynomial as:
`(x-1-3i)*(x-1+3i)`.