Find the real zeros of P(x)=x^4-7x^3+14x^2-3x-9?

Question

The answers are 3, 1 plus or minus the square root of 5 over 2. I factored it into (x-3)(X^3-4x^2+2x+3) but having trouble factoriing the 2nd factor. Please explain your answer because I will die if you don't help me!

Answer 1

Factor x³-4x²+2x+3 into (x-3)(x²-x-1) - thus you have (x-3)(x-3)(x²-x-1). See if you don't find that more useful.

Answer 2

OK, if you don't like the idea of guessing a zero or looking at the solution like I told you earlier, I can tell you how I would improvise.
First, consider that your zero is x = a. Then, you get something like:
(x-a)(bx^2 + cx + d) = x^3 - 4x^2 + 2x + 3 = 0
Even if you factor that, I think you still have to make a guess, but you have a guideline for your guess. a and d are the only terms of degree zero (constants) in the expression, so you can relate them as follows: ad = 3. From that, you can say that a = 1 or 3 are the most likely answers. You can then divide with synthetic division and get a quadratic that shares the other two zeros with your original polynomial.

Note that factoring works in this case if you make the assumption that your zero is an integer. However, there are infinitely many more equations that cannot be easily factored than can be easily factored. Guessing integers as your factors will probably get you through the class because textbooks for introductory courses do not usually use realistic numbers in problems, but it is often a huge waste of time.

Answer 3

Alright.. i don't know if this will help. but i went thru the problem, or tried to. the answers that you got weren't the answer's that i got, however. so if my work is incorrect, i'm sorry. here's what i did.

(x-3)(x^3-4x^2+2x+3)=0 i set it =0 because you get real 0's that way

(x-3)=0, then x=3

(x^3-4x^2+2x+3)=0
x^3-4x^2+2x=-3 subtract 3 from both sides
x(x^2-4x+2)=-3 factor out an x, x=-3

x^2-4x+2=-3
x^2-4x+5=0 add 3 to both sides
x=(4± (square root of) (-4)^2-4(1)(5))/2 Quadratic formula

QF: x= -b ± square root of (b^2)-4ac all divided by 2a

with that i got x=2 ± 2i, however, the question asked for real zero's, so those do not work.

The real zero's that i got for this problem are x=3 and x=-3. when checked, however, x=-3 was found to be an extraneious solution, so the only real zero I found was x=3.

i hope this helps!!

Answer 4

sorry i can't help you solve it, but here is a site that will answer it for you also factor it for you if it can be factored.

www.quickmath.com

For ex:, your problem factors into (x - 3)(x - 3)(x^2 - x - 1)

x^2 - x - 1
x = (-b ± sqrt(b^2 - 4ac))/2a

x = (-(-1) ± sqrt((-1)^2 - 4(1)(-1)))/(2(1))
x = (1 ± sqrt(1 + 4))/2
x = (1 ± sqrt(5))/2
x = (1/2)(1 ± sqrt(5))

so answers are x = 3, (1/2)(1 + sqrt(5)), or (1/2)(1 - sqrt(5))

Answer 5

you've already answered your own question dear, 3 occurs twice.

fully factored your equation is (x-3)(x-3)(x^2-x-1). long division using X-3 twice. The last part can be found using quadratic formula.