Use the given zero to find remaining zeros. Show work.
h(x)=x^4-12x^3+28x^2+196x-1845 zero: 4-5i

Question

Use the given zero to find remaining zeros. Show work. 
h(x)=x^4-12x^3+28x^2+196x-1845 zero: 4-5i

anonymous · Answer

Okay ill give it a stap.. So they told you that 4-5i was a root, the funny thing about complex/imaginary roots they come as pairs. More formally called conjugate pairs. I claim that an other root is 4+5i since it will complete the pair. 
$$x^4-12x^3+28x^2+196x-1845=(x-4-5i)(x+4-5i)(ax^2+bx+c)$$
$$= ax^4+(b-8a)x^3+(41a-8b+c)x^2+(41b-8c)x+41c$$
Now set the coefficients of the orginal equation to the new one interms of a,b,c.
 $$a=1$$
$$b-8a= -12$$
$$41a-8b+c=28$$
$$41b-8c=196$$
$$41c =- 1845$$
Solve for a,b,c and then all you need to do is find the roots of $$ax^2+bx+c.$$
which will be $$x=-5 $$and $$x= 9$$