I have some questions I'm not sure about:

1) Given that 2x³+ax²+x-12 leaves a remainder of 6 when divided by x+2, find the value of the constant 1. Hence solve the equation 2x³+ax²+x-12=0.

2) f(x)&#8801;2x³+px²+qx+6 where p and q are constants. When f(x) is divided by (x+1), the remainder is 12. When f(x) is divided by (x-1), the remainder is -6.
a) Find the value of p and the value of q.
b) Show that f(1/2)=0 and hence write f(x) as the product of three linear factors.

3) Given that (x-2) is a factor of f(x), where f(x) &#8801;x³-x²+Ax+B, find an equation satisfied by the constants A and B. Given, further, that when f(x) is divided by (x-3) the remainder is 10, find a second equation satisfied by A and B. Solve your equations to find A and B. Using your values of A and B, find 3 values of x for which f(x)=0.
no. 1 as follows,

if x +2 leaves 6 then :

2(-2)^3 + a(-2)^2 + (-2) - 12 = 6
4a = 36
a = 9
2x^3 + 9x^2 + x - 12 = 0 (used bit of guess work here, put x = 1, came to 0, therefore x - 1 is factor)

divide 2x^3 + 9x^2 + x - 12 by (x - 1)

leaves you with 2x^2 + 11x + 12

should be ok from there on
1) Given that 2x³+ax²+x-12 leaves a remainder of 6 when divided by x+2, find the value of the constant 1. Hence solve the equation 2x³+ax²+x-12=0.

I'll assume you mean 'the constant a'.
What does the information given tell you about the value of the expression at a particular value of x? Can you use this fact to form an equation that you can solve to find a?
For the second part, you'll need to use trial and error to find a value of x which satisfies the equation and then use long division to form an equation (linear)(quadratic)=0 which implies either linear=0 or quadratic=0, both of which are easy to solve.

2) f(x)&#8801;2x³+px²+qx+6 where p and q are constants. When f(x) is divided by (x+1), the remainder is 12. When f(x) is divided by (x-1), the remainder is -6.
a) Find the value of p and the value of q.
b) Show that f(1/2)=0 and hence write f(x) as the product of three linear factors.

Part (a) requires a similar idea to that used in the first question albeit slightly extended - try to use the information you have (however confusing it might seem at first).
Part (b) enables you to use the factor theorem to write f(x) as (linear)(quadratic) and you can then factorise the quadratic to write it as three linear factors.

3) Given that (x-2) is a factor of f(x), where f(x) &#8801;x³-x²+Ax+B, find an equation satisfied by the constants A and B. Given, further, that when f(x) is divided by (x-3) the remainder is 10, find a second equation satisfied by A and B. Solve your equations to find A and B. Using your values of A and B, find 3 values of x for which f(x)=0.

Similar to the above.
turtle2
I have some questions I'm not sure about:

1) Given that 2x³+ax²+x-12 leaves a remainder of 6 when divided by x+2, find the value of the constant 1. Hence solve the equation 2x³+ax²+x-12=0.

2) f(x)&#8801;2x³+px²+qx+6 where p and q are constants. When f(x) is divided by (x+1), the remainder is 12. When f(x) is divided by (x-1), the remainder is -6.
a) Find the value of p and the value of q.
b) Show that f(1/2)=0 and hence write f(x) as the product of three linear factors.

3) Given that (x-2) is a factor of f(x), where f(x) &#8801;x³-x²+Ax+B, find an equation satisfied by the constants A and B. Given, further, that when f(x) is divided by (x-3) the remainder is 10, find a second equation satisfied by A and B. Solve your equations to find A and B. Using your values of A and B, find 3 values of x for which f(x)=0.

2)a) f(x)=2x3 +px2 + qx + 6

f(-1)=12
f(1)=-6

f(-1)=-2+p-q+6=12
f(1)= 2+p+q+6=-6

p-q=8
p+q=-14

2p=-6
p=-3 and q=-11
turtle2
I have some questions I'm not sure about:

1) Given that 2x³+ax²+x-12 leaves a remainder of 6 when divided by x+2, find the value of the constant 1. Hence solve the equation 2x³+ax²+x-12=0.

2) f(x)&#8801;2x³+px²+qx+6 where p and q are constants. When f(x) is divided by (x+1), the remainder is 12. When f(x) is divided by (x-1), the remainder is -6.
a) Find the value of p and the value of q.
b) Show that f(1/2)=0 and hence write f(x) as the product of three linear factors.

3) Given that (x-2) is a factor of f(x), where f(x) &#8801;x³-x²+Ax+B, find an equation satisfied by the constants A and B. Given, further, that when f(x) is divided by (x-3) the remainder is 10, find a second equation satisfied by A and B. Solve your equations to find A and B. Using your values of A and B, find 3 values of x for which f(x)=0.

for 2b use either algebraic long division, or substitute the value of x into the equation of f(x) to prove f(1/2)=0