Answer:
If [tex]h(x) = x^2 + 2x + 10[/tex] then [tex](x+1-3i)*(x+1+3i)[/tex].
If [tex]h(x) = x^2 - 2x + 10[/tex] then [tex](x-1-3i)*(x-1+3i)[/tex].
Step-by-step 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:
[tex]ax^2+bx+c[/tex]
Because the given polynomial expression is a quadratic equation, we can use the following equations for calculating the roots:
[tex]x1=(-b/2a)+\sqrt{b^2-4ac}/2a[/tex]
[tex]x1=(-b/2a)-\sqrt{b^2-4ac}/2a[/tex]
Since the second term sign is not given, then we can write the expression as:
[tex]x^2+s2x+10[/tex], in which a=1, b=s2 where 's' represents a sign (- or +), and c=10.
Using the equation for finding the roots we obtain:
[tex]x1=(-sb/2a)+\sqrt{b^2-4ac}/2a[/tex]
[tex]x1=(-s2/2*1)+\sqrt{2^2-4*1*10}/(2*1)[/tex]; notice that [tex](sb)^{2} = 2^{2}[/tex]
[tex]x1=(-s1)+\sqrt{-36}/2[/tex]
[tex]x1=-(s1)+6i/2[/tex]
[tex]x1=-(s1)+3i[/tex]
[tex]x2=(-sb/2a)-\sqrt{b^2-4ac}/2a[/tex]
[tex]x2=(-s2/2*1)-\sqrt{2^2-4*1*10}/(2*1)[/tex]; notice that (s2)²=2^2
[tex]x2=(-s1)-\sqrt{-36}/2[/tex]
[tex]x2=-(s1)-6i/2[/tex]
[tex]x2=-(s1)-3i[/tex]
If we consider 's' as possitive (+) the roots are:
[tex]x1=-1+3i[/tex] and [tex]x2=-1-3i[/tex]
Whereas if we consider 's' as negative (-) the roots are:
[tex]x1=1+3i[/tex] and [tex]x2=1-3i[/tex]
The above means that if the equation is [tex]h(x) = x^2 + 2x + 10[/tex], then we can express the polynomial as: [tex](x+1-3i)*(x+1+3i)[/tex].
But, if the equation is [tex]h(x) = x^2 - 2x + 10[/tex], then we can express the polynomial as: [tex](x-1-3i)*(x-1+3i)[/tex].
