Solutions Polynomials
1a. 
First find the
factors of 6 i.e. 
y(1) = 1 4 + 1 + 6 = 4
Since this remainder 
0 then by the factor theorem (x-1) is not a factor.
y(-1) = -1
4 1 + 6 = 0 
(x+1) is a factor by
the factor theorem.
Next use synthetic or long division to find the other factors.
-1 1 -4 1 6
-1 5 -1
1 -5 6
0
Factors are (x+1)(x2-5x+6) which can be further factorised as (x+1)(x-2)(x-3)
b) (x-1)(x+1)(x+2) c) (x-1)(x-2)(2x+1)
d) 
y(1) = 0 (x-1) is a factor by the factor theorem
1 1 0 -5 0 4 dont forget the 0 coefficients!
1 1 -4 -4
1 1 -4 -4 0
Factors are (x-1)(x3 + x2 4x 4) repeat the process with the cubic until fully factorised.
= (x-1)(x-2)(x+1)(x+2)
2. 
= (x-1)(x-2)(x+2) So
a is the point (-2,0), b(1,0), c(2,0) and d(0,4)
3.
y = 2x3
+ px2 7x 12
-1 2 p -7 -12
-2 -(p-2) p+5
2 (p-2) -p-5 p - 7 = 0 the remainder must = 0 since
(x+1)
is a factor
So p = 7
Factors are (x+1)(2x2+5x-12)
(x+1)(2x-3)(x+4)
4. y = 4x3 + px2 11x + 6 p = -4 y = (x-2)(2x+3)(2x-1)
5. y = 2x3 + px2 + qx + 6 p = -3, q = -11 y = (x+2)(x-3)(2x-1)
6. y = 2x3 - 3x2 + x + 1 and the line y = 4x 1
First let 2x3 - 3x2 + x + 1 = 4x 1
2x3 3x2 3x + 2 = 0 now factorise this polynomial
Find the factors of 2 i.e. 
let the poly. be
f(x) Find a such that f(a) = 0 and (x-a)
is a factor.
Factors are (2x-1)(x+1)(x-2) solutions at 
, x = -1, x = 2
To find the y coordinates substitute these values into y = 4x - 1
When y = 1 (
,1)
when x = -1 y = -5 (-1,-5)
when x = 2 y = 7 (2,7) this shows that the line crosses the curve at these three points.