-17
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You would use the quadratic formula. x = -b +/- √b^2 - 4ac / 2a
x^2 - 4x - 21 = 0
a=1
b= -4
c= -21
Plug it into the quadratic equation and solve. :)
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10
(2x + 4)(2x - 4) to check your answer, use the FOIL method (first, outside, inside, last) F.... 2x x 2x = 4x^2 0.... 2x x -4 = -8x I..... 2x x 4 = 8x L.... 4 x -4 = -16 so 4x^2 - 16 = (2x + 4)(2x - 4)
Dividend: x3+4x2-9x-36 Divisor: x+3 Quotient: x2+x-12
You would probably use a power-reduction trig identity to solve this equation. This states that sin2(x) = (1 - cos(2x))/2 Therefore, sin2(2x) = (1 - cos(4x))/2, or (1/2)(1 - cos(4x)) So, ∫ (1/2)(1 - cos(4x)) dx = (1/2) ∫ (1 - cos(4x)) dx. Then, ∫ (1-cos4x)dx = x - (1/4)sin(4x) + c Now, multiply that by (1/2) to get: (x - (sin(4x)/4) + c)/2 Since c is an arbitrary constant, we have: ½(x - sin(4x) / 4) + c OR 1/8 * (4x - sin(4x)) + c