Here is the final algebra problem..?

Question

Divide:

(4b^3+20b^2+19b-7) / (b+4)

Please explain!!

Answer 1

4b^2+4b+3
___________________
b+4|4b^3+20b^2+19b-7
- 4b^3+16b^2
4b^2+19b
- 4b^2+16b
3b-7
- 3b+12
-19

4b^2+4b+3+-19/(b+4)

Answer 2

the area of a triangle is declared as: A = a million/2bh, the place b = length of the backside and h = height. the area of a circle is declared as: A = pie(r)^2 = 3.14(r)^2, the place pie = 3.14 and r = radius. in view that your situation stated that the backside of the triangle = 2x and the peak = x; then the area = a million/2(2x)(x) = a million/2(2x^2) = 2x^2/2 = x^2 the area of the circle = 3.14(3x)^2 a. the area of the circle no longer lined with the help of the triangle is 3.14(3x)^2 - x^2 b. the area whilst x = 4in. this area is the area of the circle minus the area of the triangle. 3.14(3x)^2 - x^2 3.14[(3)(4)]^2 - 4^2 3.14(12)^2 - sixteen 3.14(one hundred forty four) - sixteen 452.sixteen - sixteen 436.sixteen sq. inches is the area whilst x = 4in. you could tutor your answer with the help of adjusting for the two areas and then subtracting the area of the triangle from the area of the circle and you gets the area of the the circle no longer lined with the help of the triangle. area of the triangle = 3.14[(3)(4)]^2 = 3.14(12)^2 = 3.14(one hundred forty four) = 452.sixteen area of the triangle = a million/2(2)(4)(4) = a million/2(32) = sixteen

Answer 3

divide it just like arithmetic