Question

Find all the zeros of the function and write the polynomial as a product of linear factors:
f(x) = x^2 - x + 56

Answer 1

ax^2 + bx + c format. x = [-b +-sqrt(b^2 - 4ac)] / (2a). The determinant is b^2 - 4ac. If >= 0, then there are real zeros. If not, only complex zeros.

Answer 2

so i have to put it in quadratic formula?

Answer 3

or completing the square?

Answer 4

Short-cut is just to look at the determinant, b^2 - 4ac. If that is <0, you don't have to go further.

Answer 5

You could complete the square, but that is a little more work.

Answer 6

pluggin in (1)^2 - 4(1)(56) sor b^2 - 4ac i got -223

Answer 7

So, no real zeros because you can't have the square root of a negative number. So, cannot be factored (unless you go into the realm of complex numbers).

Answer 8

tea the aim of this chapter is for complex numbers so im stuck on that

Answer 9

Then you have to use the full expression for the quadratic formula, x = [-b +-sqrt(b^2 - 4ac)] / (2a).

Answer 10

is 223 factorable?

Answer 11

because trying to do the quadratic formula i got stuck on that

Answer 12

If you want to see if a number is factorable, divide it successively by prime numbers until you use a prime number that when squared is greater than the number you are trying to factor. For instance, when working with 223, you can stop at trying to divide by 17 because 17^2 > 223 and you will have already exhausted all candidates for factoring.

Answer 13

So, try to factor by 2, 3, 5, 7, 11, 13, and 17.

Answer 14

To put this into complex number notation, you split the terms and use the relationship of sqrt(ab) = sqrt(a)sqrt(b) where "a" is -1.

Answer 15

And sqrt(-1) = i.