Answer:
1. y = x^2( x^-3) (x+2)
2. x=0 multiplicity 2 and x=4 multiplicity 2
3. y = 2x^3 -x^2 -50x +25
4.y =x^3 -6x^2 +5x+12
Step-by-step explanation:
1. y = x^4 - x^3 - 6x^2
Factor out an x^2
y = x^2( x^2 -x-6)
We can factor the polynomial inside the parenthesis
What multiplies to -6 and add to -1
-3*2 = -6 and -3+2 =-1
y = x^2( x^-3) (x+2)
2. f(x) = x^4 - 8x^3 + 16x^2
We need to factor out an x^3
f(x) = x^2(x^2 - 8x + 16)
We should notice inside the parentheses is the difference of two cubes
(a^2 -2ab-b^2) = (a-b)^2 where a=x and b=-4
f(x) = x^2(x-4)^2
We can use the zero product property to find the roots
0 = x^2(x-4)^2
x^2 = 0
x= 0 with multiplicity 2 ( because it is a square)
(x-4)^2 =0
x-4 =0
x=4 with multiplicity 2 ( because it is a square)
3. y = (x - 5)(x + 5)(2x - 1)
We know (x-5) (x+5) is the difference of squares
(a-b) (a+b) = a^2 -b^2 where a=x and b=5
y = (x ^2- 5^2 )(2x - 1)
y = (x ^2- 25 )(2x - 1)
Now we FOIL first outer inner last
y = x^2 * 2x + -1*x^2 + -2x*25 -1*-25
y = 2x^3 -x^2 -50x +25
Standard form is from highest exponent to lowest exponent.
4. Using the zero product property we know that
(x-a)(x-b)(x-c) =0 where a,b,c are the zero's of the function
(x--1) (x-3) (x-4) = 0
(x+1)(x-3) (x-4)
Now FOIL the first two terms
(x^2 -3x+1x-3)(x-4)
(x^2 -2x-3) (x-4)
Now we multiply x by the first term then multiply -4 by all the terms in the first term
x^2 *x -2x^2 -3x -4x^2 +8x+12
Combine like terms
x^3 -6x^2 +5x+12
This is now in standard form, since the exponents go from highest to lowest power
y =x^3 -6x^2 +5x+12
