Question

given: $x^2 - 5x + 6 = 0$
step 1: i need two numbers that multiply to 6 and add to -5. the numbers are -2 and -3.
step 2: $x^2 - 2x - 3x + 6 = 0$
step 3: $x(x - 2) - 3(x - 2) = 0$
step 4: $(x + 3)(x - 2) = 0$
step 5: $x + 3 = 0$ and $x - 2 = 0$
step 6: $x = -3$ and $x = 2$
(1 point)
\bigcirc step 1
\bigcirc step 2
\bigcirc step 3
\bigcirc step 4
\bigcirc step 5
\bigcirc step 6

Explanation:

Step1: Analyze the Equation

The given equation is \(x^2 - 5x + 6 = 0\). We need to factor it. The standard form of a quadratic equation is \(ax^2+bx + c = 0\), here \(a = 1\), \(b=-5\), \(c = 6\). We need two numbers that multiply to \(ac=6\) and add to \(b=-5\). The numbers are \(-2\) and \(-3\) as \((-2)\times(-3)=6\) and \((-2)+(-3)=-5\).

Step2: Rewrite the Middle Term

Using the numbers \(-2\) and \(-3\), we rewrite the middle term of the quadratic equation: \(x^2-2x - 3x + 6 = 0\). This step is about splitting the linear term into two terms using the factors we found.

Step3: Factor by Grouping

Group the first two terms and the last two terms: \(x(x - 2)-3(x - 2)=0\). Here we factor out the greatest common factor from each group.

Step4: Factor Out the Common Binomial

Now, we can factor out the common binomial \((x - 2)\) from both terms: \((x - 3)(x - 2)=0\). Wait, there is a mistake here. Wait, in step 3 we have \(x(x - 2)-3(x - 2)=0\), factoring out \((x - 2)\) gives \((x - 2)(x - 3)=0\), but in the given step 4 it is \((x + 3)(x - 2)=0\) which is incorrect. So the error is in step 4. Let's check the options, the question is probably asking which step has an error. Let's re - examine the steps:

Step 1: Correct (finding numbers that multiply to 6 and add to - 5: - 2 and - 3)

Step 2: Correct (rewriting the middle term: \(x^2-2x-3x + 6 = 0\))

Step 3: Correct (factoring by grouping: \(x(x - 2)-3(x - 2)=0\))

Step 4: Incorrect. Because \(x(x - 2)-3(x - 2)=(x - 2)(x - 3)\) not \((x + 3)(x - 2)\)

Step 5: If step 4 is wrong, step 5 which is \(x + 3 = 0\) and \(x - 2 = 0\) is also wrong (it should be \(x-3 = 0\) and \(x - 2 = 0\))

Step 6: If step 5 is wrong, step 6 is wrong (it should be \(x = 3\) and \(x = 2\) not \(x=-3\) and \(x = 2\))

But the question is about which step has an error. From the given steps, step 4 has an incorrect factorization. So the answer is Step 4.

Answer:

Step 4