Answer:
3 ( x-1) (x+1) is the factored form
zeros are x=1, -1
Step-by-step explanation:
3x^2-3
Factor out a 3
3 ( x^2 -1)
What is inside the parentheses is the difference of squares
( a^2 -b^2) = ( a-b)(a+b)
3 ( x-1) (x+1)
To find the zeros, set this equal to zero
3 ( x-1) (x+1) =0
Using the zero product property
x-1 =0 x+1 =0
x=1, x=-1
23+8X is the answer to this problem
The unknown shape could be a rectangle, parallelogram, or kite.
:)
p-6p+7=3(2p-3)-4(-10+4p
We move all terms to the left:
p-6p+7-(3(2p-3)-4(-10+4p)=0
We add all the numbers together, and all the variables
p-6p-(3(2p-3)-4(4p-10)+7=0
We add all the numbers together, and all the variables
-5p-(3(2p-3)-4(4p-10)+7=0
Answer:
c. (x + 3)
Step-by-step explanation:
using factor theorem
if x - 3 is a factor then p(a) = 0
p(a)= x^3 - 3x^2 - 4x + 12
a.(x-3)
p(3) = (3)^3 - 3(3)^2 - 4(3) + 12
= 27 - 27 - 12 + 12
= 0
therefore x-3 is a factor
b.(x + 2)
p(-2) = (-2)^3 - 3(-2)^2 - 4(-2) + 12
= -8 -12 + 8 + 12
,= 0
therefore x + 2 is a factor
c.(x + 3)
p(-3) = (-3)^3 - 3(-3)^2 - 4(-3) + 12
= -27 -27 + 12 + 12
= -30
therefore x + 3 is not a factor
d.(x-2)
p(2) = (2)^3 - 3(2)^2 - 4(2) + 12
= 8 -12 - 8 + 12
= 0
therefore x - 2 is a factor
