$ x_1=\frac{-3-\sqrt{17}}{2},x_2=\frac{-3+\sqrt{17}}{2} $

Solve the Quadratic Equation x² + 3x + 2 = 0 by Factoring

Q: How do I know which two numbers to look for?

You need two numbers that multiply to give ac and add to give b. For $ x^2 + 3x + 2 $, find numbers that multiply to 2 and add to 3. That's 2 and 1!

Q: Why do we set each factor equal to zero?

This uses the Zero Product Property: if two things multiply to give zero, then at least one of them must be zero. So if $ (x+2)(x+1) = 0 $, then either $ x+2=0 $ or $ x+1=0 $.

Q: Can I check my factoring before solving?

Absolutely! Expand your factors: $ (x+2)(x+1) = x^2 + x + 2x + 2 = x^2 + 3x + 2 $. If you get back to the original equation, your factoring is correct!

Q: What if a doesn't equal 1?

When $ a \neq 1 $, factoring gets trickier! You still look for numbers that multiply to ac and add to b, but the factoring pattern changes. Consider using the quadratic formula for complex cases.

Quadratic Factoring with Linear Factors

$x^2+3x+2=0$

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Step-by-step video solution

Watch the teacher solve the problem with clear explanations

00:00 Find X

00:03 Let's look at the coefficients (each number is actually multiplied by 1)

00:10 Let's use the roots formula to find the possible solutions

00:17 Let's substitute appropriate values and solve to find the solutions

00:28 Let's calculate the square and products

00:33 The square root of 1 equals 1

00:36 These are the 2 options (addition and subtraction)

00:42 And this is the solution to the question

Step-by-step written solution

Follow each step carefully to understand the complete solution

Understand the problem

$x^2+3x+2=0$

Step-by-step solution

To solve the quadratic equation $x^2 + 3x + 2 = 0$ , we proceed as follows:

First, we identify that the equation is in standard form, where:

$a = 1$
$b = 3$
$c = 2$

Next, we attempt to factor the quadratic equation. We look for two numbers that multiply to $ac = 1 \times 2 = 2$ and add up to $b = 3$ . The numbers $2$ and $1$ satisfy these conditions because $2 \times 1 = 2$ and $2 + 1 = 3$ .

Thus, we can factor the equation as:

$x^2 + 3x + 2 = (x + 2)(x + 1) = 0$

Now, we solve for $x$ by setting each factor equal to zero:

$x + 2 = 0 \rightarrow x = -2$
$x + 1 = 0 \rightarrow x = -1$

Therefore, the solutions to the quadratic equation are $x_1 = -2$ and $x_2 = -1$ .

Since the problem is multiple-choice and we need to confirm the correct answer, we compare these solutions with the given choices. The correct choice is:

: $x_1=-2, x_2=-1$

Thus, the correct answer to the problem is:

\ $x_1 = -2, x_2 = -1$

Final Answer

$x_1=-2,x_2=-1$

Key Points to Remember

Essential concepts to master this topic

Factoring Rule: Find two numbers that multiply to ac and add to b
Factor Pattern: $x^2 + 3x + 2 = (x + 2)(x + 1)$ using numbers 2 and 1
Solution Check: Verify $(-2)^2 + 3(-2) + 2 = 0$ and $(-1)^2 + 3(-1) + 2 = 0$ ✓

Common Mistakes

Avoid these frequent errors

Finding numbers that multiply to c instead of ac
Don't look for numbers that multiply to just c = 2! This ignores the coefficient a and leads to wrong factorization. When a = 1, you need numbers that multiply to ac = 1×2 = 2, but always consider the full product ac. Always multiply a times c first, then find your factor pairs.

Practice Quiz

Test your knowledge with interactive questions

a = coefficient of x²

b = coefficient of x

c = coefficient of the constant term

What is the value of $$ c $$ in the function $$ y=-x^2+25x $$ ?

FAQ

Everything you need to know about this question

What if the quadratic doesn't factor easily?

Not all quadratics factor with nice integer solutions! If you can't find two integers that work, try the quadratic formula instead: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

How do I know which two numbers to look for?

You need two numbers that multiply to give ac and add to give b. For $x^2 + 3x + 2$ , find numbers that multiply to 2 and add to 3. That's 2 and 1!

Why do we set each factor equal to zero?

This uses the Zero Product Property: if two things multiply to give zero, then at least one of them must be zero. So if $(x+2)(x+1) = 0$ , then either $x+2=0$ or $x+1=0$ .

Can I check my factoring before solving?

Absolutely! Expand your factors: $(x+2)(x+1) = x^2 + x + 2x + 2 = x^2 + 3x + 2$ . If you get back to the original equation, your factoring is correct!

What if a doesn't equal 1?

When $a \neq 1$ , factoring gets trickier! You still look for numbers that multiply to ac and add to b, but the factoring pattern changes. Consider using the quadratic formula for complex cases.

Do the solutions always come out negative?

Not at all! The signs of your solutions depend on the specific equation. In this case, both solutions are negative because we factored as $(x+2)(x+1)$ , giving us $x = -2$ and $x = -1$ .

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