Factoring with the sum/product pattern when a&b are above 10

Question

I got this curveball on Khan Academy

$$z^2−21z+68$$

$$=(z-17)(x-4)$$

It was all fine until I needed to find two numbers $a$ & $b$ such that $ab = 68$ and $a+b=-21$, and those numbers went above 10. I realised this is a huge limitation to perform factoring in general, and was wondering what techniques there might be to get around this?

For example, $ab = 748$, $a=22, b=34$
Is this kind of thing even solvable with two unknowns? Or do I need to ballpark it with a calculator?

Answer 1

You can use the quadratic formula. Any quadratic $ax^2+bx+c$ factors as $a\left(x+ \frac{b+\sqrt {b^2-4ac}}{2a}\right)\left(x+ \frac{b-\sqrt {b^2-4ac}}{2a}\right)$

Answer 2

You have to guess. But use some tricks.

$68 = 2^2*17$ so the only options are $1$ and $2^2*17=68$, or $2$ and $2*17 = 34$, or $4$ and $17$. Of those only one is in the ball park.

If you have $ab=748$ and $a+b= 56$ we have $748 = 2^2*11*17$. So there are only so many options. $11*17$ is way too big so we have to break up the $11$ and $17$. One of the terms will be $11k$ and the other will by $17j$ where $kj= 4$. $a + b = 56$ is even whereas $11$ and $17$ are odd so $j$ and $k$ must both be even so $2*11$ and $2*17$ is the only feasible option. And it .... works. $22 + 34 = 56$.

Answer 3

Consider any one factor pair of $68$, say $1 \times 68$. Now go on dividing one number by $2$ and multiplying other by $2$, this way the product remains the same but the sum changes. Do the same operation again, you get $4 \times 17$ which sums up to $21$ and you have the required factor pair.$$\begin{array}{ |p{3cm}||p{3cm}|p{3cm}|p{3cm}| } \hline a & \times & b&a+b\\\hline68 & \times & 1 &69\\ 34 &\times & 2 &36 \\ 17 &\times& 4& 21 \\\hline\end{array}$$You can go through the same procedure for $ab=748$ and $a+b=56$ although doing so may require greater practice with two-digit $ab$ as a prerequisite.

Answer 4

A few quick notes:

• 1 variable polynomials, can generalize, base 10 multiplication.
• as each digit in base ten is between 0 and 9, That's the part of the table we need to know (arguably with tricks a lot less).
• We group like terms,by the power of the variable (generalizing grouping by powers of 10)

with these three, we have enough to write 68 as 6y+8, and 21 as 2y+1 we can then check that (y+7)+4 = y+11. Since y=10 in base 10, we change 11 into y+1, grouping like terms gives us 2y+1 so they do add correctly. Checking the product, we have 4y+28, again y=10 so we have 4y+2y+8, grouping like terms gives, 6y+8 so the product also works. All this took, is keeping like terms together, and multiplication of numbers less than 10.