# 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. I've tried to solve it but my efforts are fruitless. If somebody could help to clarify the solution, I'd appreciate it!

• The right hand side should multiply to 8 since that is the constant term. As written you have $5 \times 5 = 25 \neq 8$. Note that the criss-cross method is a "guess and test" method so you should expect it to fail with certain combinations of these divisors and have to try again. For example $2 \times 4 = 1 \times 8 = 8$ so you have to choose one for the right hand side and see if it works, then try again if it fails. Hopefully they'll teach you a better method soon. Jan 31, 2018 at 13:58

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

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. • Thank you so much good sir! I thoroughly appreciate the diagram but your explanation wasn't too bad actually haha. Jan 31, 2018 at 14:14

To factor $5p^2 + 6p - 8$, we need two numbers with product $5 \cdot (-8) = -40$ and sum $6$. Consider the factorizations of $-40$. \begin{align*} -40 & = -1 \cdot 40 & -40 & = 1 \cdot (-40)\\ & = -2 \cdot 20 & & = 2 \cdot (-20)\\ & = \color{blue}{-4 \cdot 10} & & = 4 \cdot (-10)\\ & = -5 \cdot 8 & & = 5 \cdot (-8) \end{align*} Of these pairs of factors, only $-4$ and $10$ have a sum of $6$. Hence, \begin{align*} 5p^2 + 6p - 8 & = 5p^2 - 4p + 10p - 8 && \text{split the linear term}\\ & = p(5p - 4) + 2(5p - 4) && \text{factor by grouping}\\ & = (p + 2)(5p - 4) && \text{extract the common factor} \end{align*} In your criss-cross method, this amounts to matching $5p$ with $2$ to get $10p$ and $p$ with $-4$ to get $-4p$. That said, I recommend using the method outlined above to eliminate guessing.

Draw a railroad sign style $X$ like you have done. In the top triangle you will write the product $ac$ from the quadratic of the form $y=ax^2+bx+c$, then in the bottom you will write the $b$ term. You should have $-40$ on top and $6$ on the bottom. Now you need to find two numbers which add to $6$ but multiply to $-40$. You will find these numbers are $-4$ and $10$ (by listing factors of $-40$ and seeing which ones must add up to $6$). After this step we write $5p^2-4p+10p-8$. Now we must factor this expression. We have \begin{equation} 5p^2-4p+10p-8 = 5p(p+2)-4(p+2) \\ =(5p-4)(p+2). \end{equation} This is our solution.