please someone help me with my algebra homework!!?

Question

Problem 1: Given that (3x-a)(x-2)(x-7)=3x^3-32x^2+81x-70, determine the value of a. and Problem 2: Find all roots of the polynomial x^3-x^2+16x-16.

Answer 1

(3x - a)(x - 2)(x - 7) =
(3x - a)(x^2 - 7x - 2x + 14) =
(3x - a)(x^2 - 9x + 14) =
3x^3 - 27x^2 + 42x - ax^2 + 18ax - 14a
3x^3 + (-27 - a)x^2 + (42 + 18a)x - 14a

3x^3 - 32x^2 + 81

so

(-27 - a)x^2 + (42 + 18a)x - 14a = -32x^2 + 81
(-27 + 32 - a)x^2 + (42 + 18a)x + (-14a - 81) = 0
(5 - a)x^2 + (42 + 18a)x - (14a + 81) = 0
(-a + 5)x^2 + (18a + 42)x - (14a + 81) = 0
-(a - 5)x^2 + (18a + 42)x - (14a + 81) = 0

x = (-b ± sqrt(b^2 - 4ac))/(2a)

x = (-(18a + 42) ± sqrt((18a + 42)^2 - 4(-(a - 5)(-(14a + 81)))/(2(-(a - 5))

x = (-(18a + 42) ± sqrt(((18a + 42)(18a + 42)) - 4(a - 5)(14a + 81)))/(-2(a - 5))

x = (-(18a + 42) ± sqrt((324a^2 + 756a + 756a + 1764) - 4(14a^2 + 81a - 70a - 405))) / (-2(a - 5))

x = (-(18a + 42) ± sqrt(324a^2 + 1512a + 1764 - 4(14a^2 + 9a - 405))) / (-2(a - 5))

x = (-(18a + 42) ± sqrt(324a^2 + 1512a + 1764 - 56a^2 - 36a + 1620))) / (-2(a - 5))

x = (-(18a + 42) ± sqrt(268a^2 + 1476a + 3384)) / (-2(a - 5))

x = (-6(3a + 14) ± sqrt(4(67a^2 + 369a + 846)))/(-2(a - 5))

x = (-6(3a + 14) ± 2sqrt(67a^2 + 369a + 846)) / (-2(a - 5))

x = (3(3a + 14) ± sqrt(67a^2 + 369a + 846))/(a - 5)

Unless you mistyped something, 3x^3 - 32x^2 + 81 does not factor into (3x - a)(x - 2)(x - 7)

--------------------------------------------------------------------------------

x^3 - x^2 + 16x - 16
(x^3 - x^2) + (16x - 16)
x^2(x - 1) + 16(x - 1)
(x^2 + 16)(x - 1)

if you were to take this further, then

(x - 4i)(x + 4i)(x - 1)