Answer:
(2x-3) (2x+3)
zeros, x intercepts: -3/2, 3/2
Step-by-step explanation:
4x^2 -9
We know the difference of squares is a^2 -b^2
This factors into (a-b) (a+b)
Let 4x^2 =a^2
Taking the square root
2x =a
Let b^2 =9
Taking the square root
b= 3
(4x^2-9 ) = (2x-3) (2x+3)
To find the zeros, we set the equation equal to zero
(4x^2-9 ) = (2x-3) (2x+3) =0
Using the zero product property
2x-3 =0 and 2x+3 =0
2x-3+3 = 0+3 2x+3-3 = 0-3
2x=3 2x=-3
Divide by 2
2x/2 = 3/2 2x/2 = -3/2
x = 3/2 x = -3/2
These are the zeros of the equation (which are also the x intercepts)
10 :) I had this, EZ now :)
Answer:
x = 1
Step-by-step explanation:
-2x + y = 4
y = 2x + 4
2x + 3y = 20
2x + 3(2x + 4) = 20
2x + 6x + 12 = 20
8x + 12 = 20
8x = 8
x = 1
Well one third is 100
so two thirds must be 200
<span> Given polynomial x^2+8x-48 = 0</span>
<span>x^2+12x-4x-48 = 0</span>
<span>x(x+12)-4(x+12) = 0</span>
<span>(x+12)(x-4) = 0</span>
<span>x+12 = 0</span>
Subtract 12 from each side.
<span>x+12-12 = 0-12</span>
<span>x = -12</span>
<span>and x-4 = 0</span>
Add 4 to each side.
<span>x-4+4 = 0+4</span>
<span>x = 4</span>
<span>Roots are -12,4.</span>