How do you factorise $5p^2+6p-8$ using the criss-cross method? I would like to factorise the quadratic expression that is $5p^2+6p-8$ **using the criss-cross method** ; So far, it has been the only method taught to us. The criss-cross method uses the factors of the constant term **($c$)** and multiplies them by the factors of the leading coefficient **($a$)** in an "x" or "cross" shape. ![a visual representation]( I've tried to solve it but my efforts are fruitless. If somebody could help to clarify the solution, I'd appreciate it!

$$5p^2+6p-8$$

It's a bit hard to do the criss-cross method in Mathjax, so I might attempt to explain it with words. Bear with me...

**Step 1:** In a quadratic of the form $ax^2+bx+c$ we want to find $ac$. In your case, that is $5\cdot -8 = -40$.

**Step 2:** Consider the factors of $ac$ that add up to $b$. Factors of $-40$ are $\pm1,\pm2,\pm4,\pm5,\pm8,\pm10,\pm20,\pm40$. Only the pair of $-4, +10$ work. I.e. $-4+10=6$.

**Step 3:** Divide $a$ into your two factors. This step is harder to explain, but try to find factors of $a$ that divide into the factors of $ac$. In your example, $5$ divides into $10$, and $1$ divides into $4$, so what you have is $2$ and $4$. I will add a graphic at the end so you aren't lost.

**Step 4:** This is where you criss-cross. Your factors are the $p$ terms and your divided factors are the numbers. Geez, this is hard to explain. Here's a diagram.

![enter image description here](