Can someone please help? Thank you. -5) The function p(x) is defined by p(x) = x³ 3x² + 8x - 24 where xER.a)i) Find the remainder when p(x) is divided by (x - 2).ii) Find the remainder when p(x) is divided by (x-3).b) Prove that p(x) has only one real zero.c) Write down the transformation that will transform the graph ofy = p(x) onto the graph of y = 8x3-12x² + 16x - 24.

5) Given that, The function px is defined by px=x3-3x2+8x-24 where x∈ℝ. The following are determined as shown below. a) i) The remainder when px is divided by x-2 is given by Remainder=p2=23-322+82-24=8-32+16-24=-6 ii) The remainder when px is divided by x-3 is given by Remainder=p3=33-332+83-24=27-27+24-24=0 b) By factorization, we get, x3-3x2+8x-24=x-3ax2+bx+cx3-3x2+8x-24=ax3-3ax2+bx2-3bx+cx-3cx3-3x2+8x-24=ax3+-3a+bx2+-3b+cx-3c Comparing the coefficients on both sides, we get, ⇒a=1⇒-3a+b=-3⇒-3(1)+b=-3⇒b=0⇒-3b+c=8⇒-3(0)+c=8⇒c=8 Thus, px can be written as px=x-3x2+8 Let, qx=x2+8 Discriminant of the quadratic equation is ∆=02-418=-32 As the discriminant is negative, the roots of the quadratic equation are imaginary. Thus, px has only one real zero. c) The original equation is px=x3-3x2+8x-24 The transformed equation is y=8x3-12x2+16x-24 By the transformation p2x, we get, p2x=2x3-32x2+82x-24p2x=8x3-34x2+16x-24p2x=8x3-12x2+16x-24 Therefore, Required transformation is p2x