Factoring Quadratics: Step-by-Step Solutions

Hey Plastik Magazine readers! Let's break down how to factor some quadratic expressions. Factoring can seem tricky, but with a step-by-step approach, you'll get the hang of it. We're going to tackle three different quadratics today. So grab your pencils, and let's dive in!

1. Factoring $2n^2 + 3n - 9$

When we factor quadratic expressions, our primary goal is to rewrite the expression as a product of two binomials. In this case, we're dealing with $2n^2 + 3n - 9$. The approach involves a bit of trial and error combined with a systematic method. First, identify the coefficients: $a = 2$, $b = 3$, and $c = -9$. Next, calculate the product $ac$, which is $2 \times -9 = -18$. Now, we need to find two numbers that multiply to $-18$ and add up to $b = 3$. Those numbers are $6$ and $-3$ because $6 \times -3 = -18$ and $6 + (-3) = 3$.

Now, rewrite the middle term using these two numbers:

$2n^2 + 3n - 9 = 2n^2 + 6n - 3n - 9$

Next, factor by grouping. From the first two terms, $2n^2 + 6n$, we can factor out $2n$:

$2n(n + 3)$

From the last two terms, $-3n - 9$, we can factor out $-3$:

$-3(n + 3)$

Now, combine these:

$2n(n + 3) - 3(n + 3)$

Notice that $(n + 3)$ is a common factor. Factor it out:

$(n + 3)(2n - 3)$

So, the factored form of $2n^2 + 3n - 9$ is $(n + 3)(2n - 3)$. You can always check your work by expanding the factored form to see if it matches the original expression:

$(n + 3)(2n - 3) = 2n^2 - 3n + 6n - 9 = 2n^2 + 3n - 9$. It checks out!

Factoring quadratics might seem like a puzzle at first, but with practice, you'll start recognizing patterns and finding the right numbers more quickly. Keep practicing, and you'll master it in no time!

2. Factoring $5n^2 + 19n + 12$

Alright, let's move on to our next quadratic expression: $5n^2 + 19n + 12$. This one might look a bit intimidating, but don't worry, we'll tackle it together. Again, our goal is to rewrite this expression as a product of two binomials. First, identify the coefficients: $a = 5$, $b = 19$, and $c = 12$. Calculate the product $ac$, which is $5 \times 12 = 60$. We need to find two numbers that multiply to $60$ and add up to $b = 19$.

After a bit of thought, you'll find that the numbers are $15$ and $4$ because $15 \times 4 = 60$ and $15 + 4 = 19$. Now, rewrite the middle term using these two numbers:

$5n^2 + 19n + 12 = 5n^2 + 15n + 4n + 12$

Next, factor by grouping. From the first two terms, $5n^2 + 15n$, we can factor out $5n$:

$5n(n + 3)$

From the last two terms, $4n + 12$, we can factor out $4$:

$4(n + 3)$

Combine these:

$5n(n + 3) + 4(n + 3)$

Notice that $(n + 3)$ is a common factor. Factor it out:

$(n + 3)(5n + 4)$

So, the factored form of $5n^2 + 19n + 12$ is $(n + 3)(5n + 4)$. Let's check our work by expanding the factored form:

$(n + 3)(5n + 4) = 5n^2 + 4n + 15n + 12 = 5n^2 + 19n + 12$. Yep, it's correct! This method of breaking down the middle term and factoring by grouping is super useful for quadratics where the leading coefficient isn't 1. Keep this technique in your toolkit!

Remember, the key to mastering factoring quadratics is practice. The more you do it, the quicker you'll become at identifying the right numbers and applying the factoring techniques. Don't get discouraged if you don't get it right away. Each problem is a learning opportunity!

3. Factoring $9k^2 + 66k + 21$

Okay, time for our last quadratic expression: $9k^2 + 66k + 21$. At first glance, the coefficients might seem a bit large, but let's see if we can simplify things before diving into the factoring process. Notice that all the coefficients (9, 66, and 21) are divisible by 3. Let's factor out the greatest common factor, which is 3:

$9k^2 + 66k + 21 = 3(3k^2 + 22k + 7)$

Now, we need to factor the quadratic expression inside the parentheses: $3k^2 + 22k + 7$. Identify the coefficients: $a = 3$, $b = 22$, and $c = 7$. Calculate the product $ac$, which is $3 \times 7 = 21$. We need to find two numbers that multiply to $21$ and add up to $b = 22$. Those numbers are $21$ and $1$ because $21 \times 1 = 21$ and $21 + 1 = 22$.

Rewrite the middle term using these two numbers:

$3k^2 + 22k + 7 = 3k^2 + 21k + k + 7$

Next, factor by grouping. From the first two terms, $3k^2 + 21k$, we can factor out $3k$:

$3k(k + 7)$

From the last two terms, $k + 7$, we can factor out $1$:

$1(k + 7)$

Combine these:

$3k(k + 7) + 1(k + 7)$

Notice that $(k + 7)$ is a common factor. Factor it out:

$(k + 7)(3k + 1)$

Now, remember to include the factor of 3 that we factored out at the beginning:

$3(k + 7)(3k + 1)$

So, the factored form of $9k^2 + 66k + 21$ is $3(k + 7)(3k + 1)$. Let's check our work by expanding the factored form:

$3(k + 7)(3k + 1) = 3(3k^2 + k + 21k + 7) = 3(3k^2 + 22k + 7) = 9k^2 + 66k + 21$. Perfect, it matches the original expression!

Factoring out the greatest common factor first made this problem much easier to handle. Always look for opportunities to simplify before diving into the more complex factoring techniques. This can save you time and reduce the chances of making mistakes.

Conclusion

And there you have it, folks! We've factored three different quadratic expressions step by step. Remember, factoring is all about practice, patience, and a systematic approach. Whether you're prepping for an exam or just brushing up on your algebra skills, these techniques will come in handy. Keep practicing, and you'll become a factoring pro in no time!