If the remainder on division of x^3+2x^2+kx+3 by x-3 is 21,find the quotient and the value of k. hence, find the zeroes of the cubic polynomial x^3+2x^2+kx-18.

Let p(x)=x^3+2x^2+kx+3 
On dividing p(x) by x-3, the remainder is 21. Therefore,
 P(3)=21
 Substituting x=3 in p(x)
 P(3)=3^3 +2*(3)^2+k*3+3
  =27+18+3k+3
  =48+3k
 We know that, p(3)=21.
 So, 48+3k=21
  3k=21-48
  3k=-27
  k=-27/3 =-9
 now, p(x) =x^3+2x^2-9x-18 
 -2 is a factor of p(x) on inspection. Therefore, divide p(x) by x+2 to find the
zeroes of the polynomial.
 On dividing, we get the factors to be, (x^2-9)(x+2)
 (x^2-3^2)(x+2)
 Factorizing using the identity a^2-b^2=(a+b)(a-b) we get,
 (x+3)(x-3)(x+2)
 Therefore, the zeroes of the polynomials are -3,+3 and -2.

The correct answer is -20y

Intersecting secant-tangent theorem states that if a tangent segment and a secant segment are drawn to a circle from an exterior point, then the square of the measure of the tangent segment is equal to the product of the measures of the secant segment and its external secant segment.so to break it down: 1. pq acts as the tangent and ps as the secant with an intersection at p.sr = 21rp = 3x + 3pq = 4x + 4following to equation: tangent squared = the two segments multiplied by each otherso (4x + 4) ^ 2 = (3x + 3) • (21) simplifies to 16x^2 - 31x - 47 = 0 or (16x - 47)(x + 1) x = -1 or x = 47/162. pq acts as the tangent and ps as the secant with an intersection at p (again! )sr = 16rp = xpq = 15(15) ^ 2 = 16 • x225 = 16xx = 225/163. the interesting chords are proportional to each other so a ratio is possible to set up: ap dp 10 3 + x = = —— = pc pb 8 xcross multiple to get 10x = (3 + x)(8)x = 12