Question

factorise: 4x4 -x2 -12x -36

Answer 1

P(x) = 4 x⁴ - x² - 12 x - 36
By checking for x = 1, 2, 3. we find that x = 2 gives 0 as the value.
So (x-2) is a factor.

Also from the change of signs of coefficients we know there is at most one positive root. 

P(x) = (x-2)(4 x³ + a x² + b x + 18)
Expanding RHS and comparing coefficients we find that a = 8. b = 15.

Now Q(x) = 4x³ + 8 x² + 15 x + 18
There are possibly some negative roots and some imaginary roots too.

  We can use substitution of x = y - 2/3.  Then the x² term disappears. Then we get:
            Q9x) = Q(y) = y³ + 29/12 y + 70/27 

Now substitute y = t - 29/(36*t)
       we get t⁶ + 70/27 t³ - 29³/36³ = 0
 
    Solving it we get  t³ and then t, then find y = -5/6.  So x = -3/2.  The other values are imaginary.


So given polynomial P(x) = (x - 2)(2x+3) (2x²+ax+6)
         Find a by expanding the terms.  a = 1.
Answer is that. as the quadratic does not have real roots.