Matching Quadratic Expressions With Their Factors

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Matching Quadratic Expressions with Their Factors

Hey guys! Today, we're diving into the exciting world of quadratic expressions and their factors. Factoring quadratics is a fundamental skill in algebra, and it's super useful for solving equations, simplifying expressions, and understanding the behavior of polynomial functions. In this article, we'll break down how to match each quadratic expression with its correct factors. Let's get started and make sure we get this down pat!

I) Factoring 2x2+5x−32x^2 + 5x - 3

Let's tackle the first quadratic expression: 2x2+5x−32x^2 + 5x - 3. To factor this, we need to find two binomials that, when multiplied together, give us this exact quadratic. We're looking for something in the form (ax + b)(cx + d) where a, b, c, and d are constants. A common technique is to look for two numbers that multiply to give us the product of the leading coefficient (2) and the constant term (-3), which is -6, and that add up to the middle coefficient, which is 5. The numbers that fit this description are 6 and -1.

So, we can rewrite the middle term using these numbers: 2x2+6x−x−32x^2 + 6x - x - 3. Now, we factor by grouping. From the first two terms, 2x2+6x2x^2 + 6x, we can factor out a 2x2x, giving us 2x(x+3)2x(x + 3). From the last two terms, −x−3-x - 3, we can factor out a -1, giving us −1(x+3)-1(x + 3). Now, we have 2x(x+3)−1(x+3)2x(x + 3) - 1(x + 3). Notice that (x+3)(x + 3) is a common factor. Factoring that out, we get (2x−1)(x+3)(2x - 1)(x + 3).

Looking at our factor choices, we see that (2x−1)(2x - 1) corresponds to option F and (x+3)(x + 3) corresponds to option D. Therefore, the factors of 2x2+5x−32x^2 + 5x - 3 are (2x−1)(2x - 1) and (x+3)(x + 3). So, I) matches with D and F. We can quickly verify this by expanding (2x−1)(x+3)(2x - 1)(x + 3):

(2x−1)(x+3)=2x(x)+2x(3)−1(x)−1(3)=2x2+6x−x−3=2x2+5x−3(2x - 1)(x + 3) = 2x(x) + 2x(3) - 1(x) - 1(3) = 2x^2 + 6x - x - 3 = 2x^2 + 5x - 3. This confirms our factoring is correct. Remember, practice makes perfect, and understanding the underlying logic will help you nail these every time!

II) Factoring 2x2−5x−32x^2 - 5x - 3

Now, let's move on to the second quadratic expression: 2x2−5x−32x^2 - 5x - 3. This looks pretty similar to the first one, but notice the sign change in the middle term. This small change can make a big difference in the factors! We're still looking for two numbers that multiply to give us the product of the leading coefficient (2) and the constant term (-3), which is -6, but this time they need to add up to -5. The numbers that satisfy these conditions are -6 and 1.

We rewrite the middle term using these numbers: 2x2−6x+x−32x^2 - 6x + x - 3. Factoring by grouping, we factor out a 2x2x from the first two terms, 2x2−6x2x^2 - 6x, giving us 2x(x−3)2x(x - 3). From the last two terms, x−3x - 3, we can factor out a 1, giving us 1(x−3)1(x - 3). Now, we have 2x(x−3)+1(x−3)2x(x - 3) + 1(x - 3). Notice that (x−3)(x - 3) is a common factor. Factoring that out, we get (2x+1)(x−3)(2x + 1)(x - 3).

Looking at our factor choices, we see that (2x+1)(2x + 1) corresponds to option E and (x−3)(x - 3) corresponds to option B. Therefore, the factors of 2x2−5x−32x^2 - 5x - 3 are (2x+1)(2x + 1) and (x−3)(x - 3). So, II) matches with B and E. Let's verify this:

(2x+1)(x−3)=2x(x)+2x(−3)+1(x)+1(−3)=2x2−6x+x−3=2x2−5x−3(2x + 1)(x - 3) = 2x(x) + 2x(-3) + 1(x) + 1(-3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3. Again, our factoring is correct. It's all about paying attention to those signs and finding the right combination!

III) Factoring 2x2+5x+32x^2 + 5x + 3

Finally, let's factor the third quadratic expression: 2x2+5x+32x^2 + 5x + 3. In this case, all the terms are positive, which might make it a bit easier to factor. We're looking for two numbers that multiply to give us the product of the leading coefficient (2) and the constant term (3), which is 6, and that add up to the middle coefficient, which is 5. The numbers that fit this description are 2 and 3.

We rewrite the middle term using these numbers: 2x2+2x+3x+32x^2 + 2x + 3x + 3. Factoring by grouping, we factor out a 2x2x from the first two terms, 2x2+2x2x^2 + 2x, giving us 2x(x+1)2x(x + 1). From the last two terms, 3x+33x + 3, we can factor out a 3, giving us 3(x+1)3(x + 1). Now, we have 2x(x+1)+3(x+1)2x(x + 1) + 3(x + 1). Notice that (x+1)(x + 1) is a common factor. Factoring that out, we get (2x+3)(x+1)(2x + 3)(x + 1).

Looking at our factor choices, we see that (2x+3)(2x + 3) corresponds to option C and (x+1)(x + 1) corresponds to option A. Therefore, the factors of 2x2+5x+32x^2 + 5x + 3 are (2x+3)(2x + 3) and (x+1)(x + 1). So, III) matches with A and C. Let's check our work:

(2x+3)(x+1)=2x(x)+2x(1)+3(x)+3(1)=2x2+2x+3x+3=2x2+5x+3(2x + 3)(x + 1) = 2x(x) + 2x(1) + 3(x) + 3(1) = 2x^2 + 2x + 3x + 3 = 2x^2 + 5x + 3. Our factoring is correct once again! By now, you should be getting pretty comfortable with this process.

Summary of Matches

Okay, let's summarize our findings:

  • I) 2x2+5x−32x^2 + 5x - 3 matches with D) (x+3)(x + 3) and F) (2x−1)(2x - 1)
  • II) 2x2−5x−32x^2 - 5x - 3 matches with B) (x−3)(x - 3) and E) (2x+1)(2x + 1)
  • III) 2x2+5x+32x^2 + 5x + 3 matches with A) (x+1)(x + 1) and C) (2x+3)(2x + 3)

Conclusion

And there you have it! We've successfully matched each quadratic expression with its factors. Factoring quadratics can seem tricky at first, but with practice and a solid understanding of the underlying principles, you'll become a pro in no time. Keep practicing, and don't be afraid to ask for help when you need it. You got this! Keep up the great work, and remember to always double-check your answers. Happy factoring!