Question

Write the polynomial as the product of linear factors. h(x) = x^2 ? 2x + 10

Answer 1

Answer:

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

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

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:

$ax^2+bx+c$

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

$x1=(-b/2a)+\sqrt{b^2-4ac}/2a$

$x1=(-b/2a)-\sqrt{b^2-4ac}/2a$

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:

$x1=(-sb/2a)+\sqrt{b^2-4ac}/2a$

$x1=(-s2/2*1)+\sqrt{2^2-4*1*10}/(2*1)$; notice that $(sb)^{2} = 2^{2}$

$x1=(-s1)+\sqrt{-36}/2$

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

$x1=-(s1)+3i$

$x2=(-sb/2a)-\sqrt{b^2-4ac}/2a$

$x2=(-s2/2*1)-\sqrt{2^2-4*1*10}/(2*1)$; notice that (s2)²=2^2

$x2=(-s1)-\sqrt{-36}/2$

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

$x2=-(s1)-3i$

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: $(x-1-3i)*(x-1+3i)$.