Factor $12q^2 - 11qr - 15r^2$

Dec 9, 2025 by Andrew McMorgan 30 views

Hey guys, welcome back to Plastik Magazine! Today, we're diving deep into the awesome world of algebra, specifically focusing on how to factor completely a quadratic expression. This skill is super fundamental, and once you nail it, you'll find that many other math problems become way easier to tackle. We're going to break down the expression $12 q^2 - 11 q r - 15 r^2$ step-by-step, so by the end of this, you'll be a factoring pro. Remember, factoring is like undoing multiplication; it's about finding the smaller expressions that multiply together to give you the original one. It’s a bit like solving a puzzle, and the satisfaction you get when you crack it is unbeatable!

Understanding Quadratic Expressions

Before we jump into factoring $12 q^2 - 11 q r - 15 r^2$ , let's quickly chat about what quadratic expressions are. Basically, a quadratic expression is a polynomial with the highest power of the variable being two. In our case, we have two variables, $q$ and $r$ , but the concept is the same. The expression $12 q^2 - 11 q r - 15 r^2$ is a quadratic trinomial because it has three terms. The terms are $12 q^2$ , $-11 q r$ , and $-15 r^2$ . When we talk about factoring these, we're looking for two binomials (expressions with two terms) that, when multiplied together, give us back our original trinomial. It sounds tricky, but there are systematic ways to approach it, ensuring you don't miss any possibilities and that you can factor completely.

The Method: Finding Two Numbers

So, how do we actually factor completely an expression like $12 q^2 - 11 q r - 15 r^2$ ? The most common method involves finding two specific numbers that help us break down the middle term. This method is often called the 'ac method' or 'splitting the middle term.' Here's the game plan:

Identify 'a', 'b', and 'c': In a standard quadratic $ax^2 + bx + c$ , $a$ is the coefficient of the squared term, $b$ is the coefficient of the linear term, and $c$ is the constant term. For our expression $12 q^2 - 11 q r - 15 r^2$ , we need to be a little careful because we have two variables. Think of $q^2$ as your $x^2$ and $r$ as a sort of unit for the $x$ term. The coefficient of the $q^2$ term (our 'a') is 12. The coefficient of the $qr$ term (our 'b') is -11. The 'constant' term, which here is $-15r^2$ , is a bit different. For the standard ac method, we'd multiply $a$ and $c$ . So, let's calculate $a imes c$ . Here, $a=12$ and the 'c' part is $-15$ . So, $a imes c = 12 imes (-15)$ . Let's do that multiplication: $12 imes 15 = 180$ . Since it's $12 imes (-15)$ , the product is $-180$ .
Find Two Numbers: Now, the crucial step is to find two numbers that multiply to give us our product ( $ac$ , which is $-180$ ) and add up to give us our middle coefficient ( $b$ , which is $-11$ ). This is where the puzzle aspect really kicks in. We need pairs of factors for $-180$ . Since the product is negative, one number must be positive, and the other must be negative. Since the sum is negative ( $-11$ ), the negative number must have a larger absolute value than the positive number.

Let's list some factor pairs of 180:

1 and 180
2 and 90
3 and 60
4 and 45
5 and 36
6 and 30
9 and 20
10 and 18
12 and 15

Now, let's consider the signs. We need one positive and one negative number that add up to $-11$ . Let's test the pairs:

$-180 + 1 = -179$
$-90 + 2 = -88$
$-60 + 3 = -57$
$-45 + 4 = -41$
$-36 + 5 = -31$
$-30 + 6 = -24$
$-20 + 9 = -11$

Boom! We found them! The two numbers are $-20$ and $9$ . They multiply to $-180$ ( $-20 imes 9 = -180$ ) and add up to $-11$ ( $-20 + 9 = -11$ ). This is a massive step towards being able to factor completely.

Splitting the Middle Term

With our magic numbers $-20$ and $9$ , we can now rewrite the middle term, $-11qr$ , as the sum of two terms using these numbers. So, $-11qr$ becomes $-20qr + 9qr$ . Our expression $12 q^2 - 11 q r - 15 r^2$ now looks like this: $12 q^2 - 20qr + 9qr - 15 r^2$ . This might seem like we're complicating things, but trust me, this is the key to unlocking the factored form. We haven't changed the value of the expression, just how it's written, which is essential for the next step in how to factor completely.

Grouping and Factoring

Now that we've split the middle term, we group the terms into pairs and factor out the greatest common factor (GCF) from each pair. Our expression is $12 q^2 - 20qr + 9qr - 15 r^2$ . Let's group the first two terms and the last two terms:

Group 1: $12 q^2 - 20qr$
Group 2: $+ 9qr - 15 r^2$

Now, find the GCF for each group:

GCF of Group 1: The GCF of $12 q^2$ and $20qr$ . Both numbers (12 and 20) are divisible by 4. Both terms have a factor of $q$ . So, the GCF is $4q$ . Factoring this out, we get: $4q(3q - 5r)$ .
GCF of Group 2: The GCF of $9qr$ and $15r^2$ . Both numbers (9 and 15) are divisible by 3. Both terms have a factor of $r$ . So, the GCF is $3r$ . Factoring this out, we get: $3r(3q - 5r)$ .

Notice something super cool here? Both resulting binomials are identical: $(3q - 5r)$ . This is exactly what we want! If the binomials inside the parentheses are the same, it means we're on the right track to factor completely. If they were different, we'd have to go back and check our steps, maybe our numbers or our factoring.

The Final Step to Factor Completely

We now have our expression in a form that looks like this: $4q(3q - 5r) + 3r(3q - 5r)$ . See that common binomial factor, $(3q - 5r)$ ? We can now factor this out just like we factored out a GCF before. Think of $(3q - 5r)$ as a single 'thing'. We have $4q$ times that 'thing', plus $3r$ times that 'thing'. So, we can factor out $(3q - 5r)$ .

What's left? We have $4q$ from the first part and $3r$ from the second part. These become the terms in our second binomial factor. So, the factored form is: $(4q + 3r)(3q - 5r)$ .

And there you have it! We have successfully factored the expression $12 q^2 - 11 q r - 15 r^2$ into $(4q + 3r)(3q - 5r)$ . This means that if you were to multiply $(4q + 3r)$ by $(3q - 5r)$ using the FOIL method (First, Outer, Inner, Last), you would get back our original expression. This confirms we have managed to factor completely.

Checking Your Work

It's always a good idea to check your answer to make sure you've factored correctly. Let's multiply our two binomials $(4q + 3r)$ and $(3q - 5r)$ to see if we get $12 q^2 - 11 q r - 15 r^2$ .

Using FOIL:

First: $(4q) imes (3q) = 12q^2$
Outer: $(4q) imes (-5r) = -20qr$
Inner: $(3r) imes (3q) = 9qr$
Last: $(3r) imes (-5r) = -15r^2$

Now, combine the terms: $12q^2 - 20qr + 9qr - 15r^2$ .

Combine the like terms (the $qr$ terms): $-20qr + 9qr = -11qr$ .

So, the result is: $12q^2 - 11qr - 15r^2$ .

And voila! It matches our original expression. This confirms that our factoring was correct and that we did indeed factor completely.

When a Polynomial is Prime

Sometimes, when you try to factor a quadratic expression, you might find that you can't break it down any further using integer coefficients. In such cases, the polynomial is called prime. For our expression $12 q^2 - 11 q r - 15 r^2$ , we successfully found two binomial factors. So, it's not prime. However, let's say you were trying to factor something like $x^2 + x + 1$ . If you tried to find two numbers that multiply to 1 and add to 1, you wouldn't find any integers that work. This is a classic example of a prime polynomial. You can't factor completely it using real numbers (unless you go into more advanced topics like complex numbers or special factoring patterns that don't apply here). It's important to recognize when a polynomial is prime because it means it's in its simplest factored form.

Conclusion

So there you have it, folks! Factoring the expression $12 q^2 - 11 q r - 15 r^2$ completely involved a systematic approach: identify $a$ , $b$ , and $c$ ; find two numbers that multiply to $ac$ and add to $b$ ; split the middle term; group terms and factor out GCFs; and finally, factor out the common binomial. This method is your go-to for tackling similar problems. Mastering how to factor completely will not only boost your confidence in algebra but also open doors to solving more complex equations and understanding functions better. Keep practicing, and you'll find these techniques become second nature. Until next time, happy factoring!