answer: y= 2x^2+2x-4
steps:
y= (2x+4)(x-1)
*FOIL method* y= (2x^2-2x+4x-4)
*simplify* y= 2x^2+2x-4
I'd definitely factor 4x^2-48 first. Factors are (4)(x^2-12).
This factors further: (4)* (x-2sqrt(3) ) * (x+2sqrt(3) )
X2 + y2 = 29
Center of the circle
x=0 and y=0
Therefore the slope of the line perpendicular to the tangent line
(y2-y1) / (x2-x1)
(5-0) / (2-0)
5/2
so the slope of the tangent
-2/5
General equation for the line
y = mx + c
if the line satisfies with the point (2,5)
5 = 2m + c
m = -2/5
5 = 2 * (-2/5) + c
c = 5 + (4/5)
c = 29/5
Therefore the equation of the line
y = mx + c
y = (-2/5) m + 29/5
689:53=13
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(x^6 - 4x^5 - 7x^3)/2x^3=x^3(x^3-4x^2-7)/2x^3= (x^3-4x^2-7)/2
