Show that x? + x + 1 is not a factor of the expression
x² - 4x + x² – 5x+6.
please explain the answer
Answers
es. They won't always format things as nicely as this. So how do we go from a regular quadratic like the above to an equation that is ready to be square-rooted?
We will have to "complete the square".
Here's how we'd have solved the last equation on the previous page, if they hadn't formatting it nicely for us.
Use completing the square to solve x2 – 4x – 8 = 0.
As noted above, this quadratic does not factor, so I can't solve the equation by factoring. And they haven't given me the equation in a form that is ready to square-root. But there is a way for me to manipulate the quadratic to put it into that ready-for-square-rooting form, so I can solve.
First, I put the loose number on the other side of the equation:
x2 – 4x – 8 = 0
x2 – 4x = 8
Then I look at the coefficient of the x-term, which is –4 in this case. I take half of this number (including the sign), which gives me –2. (I need to keep track of this value. It will simplify my work later on.)
Then I square this value to get +4, and add this squared value to both sides of the equation:
x2 – 4x + 4 = 8 + 4
x2 – 4x + 4 = 12
This process creates a quadratic expression that is a perfect square on the left-hand side of the equation. I can factor, or I can simply replace the quadratic with the squared-binomial form, which is the variable, x, together with the one-half number that I got before (and noted that I'd need later), which was –2. Either way, I get the square-rootable equation:
(x – 2)2 = 12
(I know it's a "–2" inside the parentheses because half of –4 was –2. By noting the sign when I'm finding one-half of the coefficient, I help keep myself from messing up the sign later, when I'm converting to squared-binomial form.) Read and understand