I was given algebra 2 extra credit homework over the winter break, and I need help with this polynomial. I just can't factor it. Please help, i'd appreciate it a lot, thanks.
2006-12-16 06:26:38 · 16 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
(x-1)(x-1)(x-1)
pascals triangle
or try using synthetic division
2006-12-16 06:31:09 · answer #1 · answered by Anonymous · 0⤊ 0⤋
Question Number 1 For this polynomial equation x^3 +3*x^2 -x -3 = 0, answer the following questions : A. Solve by Factorization Answer For Question 1 x^3 +3*x^2 -x -3 = 0 And we get P(x)=x^3 +3*x^2 -x -3 Now, we can look for the roots of P(x) using various Algorithm : 1A. Solve by Factorization x^3 +3*x^2 -x -3 = 0 Separate : ( x^3 -x^2 ) + ( 4*x^2 -4*x ) + ( 3*x -3 ) = 0 Commutative Law : ( x^3 +4*x^2 +3*x ) + ( -x^2 -4*x -3 ) = 0 Distributive Law : x*( x^2 +4*x +3 ) + -*( x^2 +4*x +3 ) = 0 Factor : ( x -1 )*( x^2 +4*x +3 ) = 0 Separate : ( x -1 )*( ( x^2 +x ) + ( 3*x +3 ) ) = 0 Commutative Law : ( x -1 )*( ( x^2 +3*x ) + ( x +3 ) ) = 0 Distributive Law : ( x -1 )*( x*( x +3 ) + +*( x +3 ) ) = 0 Factor : ( x -1 )*( x +1 )*( x +3 ) So the Polynomial have 3 roots : x1 = 1 x2 = -1 x3 = -3
2016-05-22 23:39:34 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Put it in your calculator.
Graph it
Where it crosses the X axis thats a zero.
Take that and use synthetic division.
If you haven't learned it though then here is a way to figure it out by using the graph.
If it crosses the X axis and "bounces back" in other words if it makes a parabola then its that answer squared. Say the graph of this equaiton makes a parabola when x = 1. Then one of the factors would be (x-1)^2. If it goes through a certain point like straight down without making a parabola then it will be that answer but not squared. So if it went through when x = 2 then a factor would be (x-2) but not squared.
Hope this helps.
2006-12-16 06:33:31 · answer #3 · answered by Ohms 2 · 0⤊ 0⤋
x^3 - 3x^2 + 3x - 1
Since you have an x^3, you know that the three factors. You know that (x - 1) ^2 = (x^2 - 2x + 1). Now multiply that by (x-1)
(x-1) (x^2 - 2x + 1) =your original equation. (x-1)^ 2 * (x -1 ) =
(x-1) (x-1) (x-1).
You can also graph the function to and see where it meets the x axis. Notice that it only ever crosses the x axis at x = 1. Since it is a cubic function, it must have three factors. Since x = 1, that means the factor at that point is (x -1). Since 1 is the only spot it ever meets the x axis, you repeat that factor 3 times, since it is a cubic function.
2006-12-16 07:05:33 · answer #4 · answered by j 4 · 0⤊ 0⤋
Factor this polynomial: x³ - 3x² + 3x - 1
Always try dividing this kind of polynomial by (x-1) first.
In this case your result will be: (x-1)(x² - 2x + 1)
which factors out to be: (x-1)(x-1)(x-1)
2006-12-16 06:35:34 · answer #5 · answered by Anonymous · 1⤊ 0⤋
You could use synthetic division , but really you should recognize that this is simply (a-b)^3 where a = x and b =1. So answer is
(x-1)^3. It pays to remember these formulae.
2006-12-16 06:56:15 · answer #6 · answered by ironduke8159 7 · 0⤊ 0⤋
Simple but effective:
(a-b)^3
(x-1)^3
2006-12-16 07:31:36 · answer #7 · answered by Bao L 3 · 0⤊ 0⤋
x-3+3x-1
2006-12-16 06:29:20 · answer #8 · answered by chicklover_563 2 · 0⤊ 0⤋
(x-1)^3 = x^3 + 3*(x)^2 *(-1) + 3* x *(-1)^2 + (-1)^3
so
x^3 - 3x^2 + 3x -1 = (x-1)^3
2006-12-16 07:01:02 · answer #9 · answered by Beta01 3 · 0⤊ 0⤋
That "man" dude is everywhere!
Can this problem be factored?
Sorry, I tried on paper. I'm just in algebra I.
2006-12-16 06:38:19 · answer #10 · answered by Life Is Great 4 · 0⤊ 0⤋