Factor 4m^2 - 4m - 5, or show that it has no integer factorisation.

Step-by-step solution

1. 1

   Try integer pairs and record the failures. With $4m^2=(2m)(2m)$ and constant $-5$, the only integer options are $(2m+1)(2m-5)$ and $(2m-1)(2m+5)$. Expanding:

   $(2m+1)(2m-5)=4m^2-8m-5,\qquad(2m-1)(2m+5)=4m^2+8m-5$

   Both give $\pm 8m$, never $-4m$, so no integer factorisation exists.

2. 2

   Confirm with the discriminant. With $a=4$, $b=-4$, $c=-5$:

   $\Delta=16+80=96$

   Since $96$ is not a perfect square ($9^2=81$, $10^2=100$), the roots are irrational and an integer factorisation was impossible from the start.

3. 3

   Find the roots. $\sqrt{96}=4\sqrt6$, so

   $m=\frac{4\pm 4\sqrt6}{8}=\frac{1\pm\sqrt6}{2}$

4. 4

   Write the factorisation over the reals. For roots $r_1,r_2$ the quadratic equals $a(m-r_1)(m-r_2)$; the leading coefficient $4$ must be carried:

   $4m^2-4m-5=4\left(m-\frac{1+\sqrt6}{2}\right)\left(m-\frac{1-\sqrt6}{2}\right)$

5. 5

   Check by expanding the root form. The product of the roots is $\tfrac{1-6}{4}=-\tfrac54=\tfrac{c}{a}$ and their sum is $1=-\tfrac{b}{a}$ $\checkmark$. Numerically the roots are $\approx 1.7247$ and $\approx-0.7247$; substituting the first gives $4(2.9746)-6.899-5\approx 0$ $\checkmark$.