Factor the quadratic 10x² − 3x − 4 into two binomials.

Step-by-step solution

1. 1

   Set up the AC product. When the leading coefficient is not $1$, trial and error over binomials is slow. Multiply the leading coefficient by the constant term instead:

   $a\cdot c=10\cdot(-4)=-40$

   You now need two integers whose product is $-40$ and whose sum is the middle coefficient $b=-3$.

2. 2

   Find the pair. Because the product is negative, the two numbers have opposite signs, and the larger magnitude must carry the minus sign to make the sum negative. Running through the factor pairs of $40$ — $(1,40),(2,20),(4,10),(5,8)$ — the last one works:

   $5\cdot(-8)=-40,\qquad 5+(-8)=-3$

3. 3

   Split the middle term using that pair. Replace $-3x$ by $5x-8x$ (the order does not matter, only the grouping that follows):

   $10x^2-3x-4=10x^2+5x-8x-4$

4. 4

   Group in pairs and factor each group.

   $(10x^2+5x)+(-8x-4)=5x(2x+1)-4(2x+1)$

   The check that you split correctly is that both groups leave the same binomial $2x+1$. If they do not match, you picked the wrong pair or grouped the wrong way.

5. 5

   Factor out the shared binomial and verify.

   $5x(2x+1)-4(2x+1)=(2x+1)(5x-4)$

   Expanding back: $(2x+1)(5x-4)=10x^2-8x+5x-4=10x^2-3x-4$, which matches. The roots are therefore $x=-\tfrac12$ and $x=\tfrac45$.