Question

Question 3 (2 points)

David's homework was to factor the polynomial completely (a) over the real numbers, (b) over the complex numbers.
x³+4x²+5x
He started :x(x² + 4x + 5) He knows that x=0 is a real root.
But there are no real factors of 5 that add up to 4???? So what does he do?

Answer 1

To factor over the complex numbers, David will need the quadratic formula.

x² + 4x + 5 is of the form ax² + bx + c where

• a = 1
• b = 4
• c = 5

Let's plug those into the quadratic formula.

`x = (-b\pm√(b^2-4ac))/(2a)\\\\x = (-4\pm√((4)^2-4(1)(5)))/(2(1))\\\\x = (-4\pm√(16-20))/(2(1))\\\\x = (-4\pm√(-4))/(2)\\\\x = (-4 \pm 2i)/(2)\\\\x = (2(-2 \pm i))/(2)\\\\x = -2 + i \text{ or } x = -2 - i\\\\`

The roots of x² + 4x + 5 are x = -2+i and x = -2-i

Now recall that if x = k is a root of a polynomial, then (x-k) is a factor of that polynomial.

For example, if x = 5 is a root then (x-5) would be a factor.

• -2+i being one root leads to the factor x-(-2+i) which simplifies to x+2-i
• -2-i being the other root leads to the factor x-(-2-i) which simplifies to x+2+i

x² + 4x + 5 factors to (x+2-i)(x+2+i)

Overall x³+4x²+5x fully factors to x(x+2-i)(x+2+i) when factoring over the complex numbers.