Looks like a pretty regular system, no parallel lines. In fact the second equation is a vertical line as written; not sure if that's a typo. Either way,

Answer: EXACTLY ONE SOLUTION

4 0

3 years ago

Find all the zeros of the polynomial function p(x) = x3 – 5x2 + 33x – 29

hram777 [196]

Answer:

$\large \boxed{\sf \ \ x=1, \ \ x=2+5i, \ \ x=2-5i \ \ }$

Step-by-step explanation:

Hello,

I assume that we are working in $\mathbb{C}$ , otherwise there is only one zero which is 1. Please consider the following.

First of all, we can notice that 1 is a trivial solution as

$p(1) = 1^3-5\cdot 1^2 + 33\cdot 1-29=1-5+33-29=0$

It means that (x-1) is a factor of p(x) so we can find two real numbers, a and b, so that we can write the following.

$p(x)=(x-1)(x^2+ax+b)=x^3+ax^2+bx-x^2-ax-b=x^3+(a-1)x^2+(b-a)x-b$

Let's identify like terms as below.

a-1 = -5 <=> a = -5 + 1 = -4

b-a = 33

-b = -29 <=> b = 29

$\boxed{ \ p(x)=(x-1)(x^2-4x+29) \ }$

Now, we need to find the zeroes of the second factor, meaning finding x so that:

$x^2-4x+29=0 \ \text{ complete the square, 29 = 25 + 4} \\ \\ x^2-2\cdot 2 \cdot x+2^2+25=0 \\ \\ (x-2)^2=-25=(5i)^2 \ \text{ take the root } \\ \\x-2=\pm 5i \ \text{ add 2 } \\ \\ x = 2+5i \ \text{ or } \ x = 2-5i$