The polynomial 4x^3 + ax + 2, where a is a constant, is denoted by p(x). It is given that 2x + 1 is a factor of p(x).

Question

(i) Find the value of a.

(ii) When a has this value

(a) factorise p(x)

(b) solve inequality p(x)> 0, justifying your answer

Answer:

I found the value of a equal to 3 by using the factor theorem.

Then I factorised p(x) by dividing p(x) by 2x+1 and got quotient 2x^2-x+2.

so p(x) = (2x+1) (2x^2-x+2)

I was not able to do number b . I dont really know what it is trying to say.

Comments

a=3.

Here’s why:

(2x+1)(2x²-x+2)=4x³-2x²+4x+2x²-x+2=4x³+3x+2.

4x³+ax+2=(2x+1)(2x²+qx+2), where q is another constant, because we know that 4x³/2x=2x², and to get the constant 2 we need a constant 2/1=2. Let’s expand the parentheses:

4x³+2qx²+4x+2x²+qx+2. There’s no x² term so q=-1 to cancel out the x² term.

Therefore qx=-x. Add this to 4x and we get 4x-x=3x. This gives us a=3.

b) When x=-½, p(x)=0, when x>-½, p(x)>0. Note that 2x²-x+2>0 always.

Answer 1

(2x+1) (2x^2-x+2)>0

First we test (2x^2-x+2) > 0

2x^2-x+2

=> 2( x^2 -x/2 +1)

=> 2( x^2 -x/2 +1 + (-1/4)^2 - (-1/4)^2)    , adding and subtracting (-1/4)^2 in order to complete the square

=> 2{(x-1/4)^2 + 15/16} and this is always greater than 0, for values of x.

Now we test (2x+1) >0

2x+1=0 => x = -1/2

So, 2x+1>0 if and only if x>-1/2

So the solution to p(x) = (2x+1) (2x^2-x+2) >0 is x>-1/2