Solving Quadratic Factoring - ©n I 2 c 001 i 2 v RKZuttyav 6 ... - Free Printable

It looks like you've uploaded a worksheet titled "Solving Quadratic Equations by Factoring", and it includes 10 quadratic equations, each with their solutions already provided in red. However, since I can't view images directly, I’ll assume you're asking me to verify the solutions or explain how to solve each equation by factoring, based on the problems listed.

Let’s go through each problem step by step, showing how to factor and solve them, and verify whether the given answers are correct.

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1) \( (x + 1)(x - 5) = 0 \)

Already factored. Set each factor equal to zero:
- \( x + 1 = 0 \Rightarrow x = -1 \)
- \( x - 5 = 0 \Rightarrow x = 5 \)

✔ Solution: \( \{-1, 5\} \) — Correct

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2) \( 3x + 8(x + 2) = 0 \)

Wait — this is not a quadratic equation. Let's simplify:

\( 3x + 8(x + 2) = 3x + 8x + 16 = 11x + 16 = 0 \)

This is linear:
\( 11x = -16 \Rightarrow x = -\frac{16}{11} \)

But the answer says \( \{-4, -2\} \), which doesn’t match.

✘ Error: This is not a quadratic equation. The given answer is incorrect.

> ⚠️ Likely typo: Perhaps the original equation was meant to be something like:
> \( 3x^2 + 8x + 2 = 0 \) or \( (3x + 8)(x + 2) = 0 \)

Let’s suppose it was:
\( (3x + 8)(x + 2) = 0 \) → then:
- \( 3x + 8 = 0 \Rightarrow x = -\frac{8}{3} \)
- \( x + 2 = 0 \Rightarrow x = -2 \)

Still not matching \( \{-4, -2\} \)

Alternatively, maybe:
\( (x + 4)(x + 2) = 0 \Rightarrow x = -4, -2 \)

So likely the intended equation was:
\( x^2 + 6x + 8 = 0 \) → factors as \( (x+4)(x+2) = 0 \)

So unless the original equation was different, the current one is not quadratic and the answer is wrong.

✔ Conclusion: Equation as written is linear, so cannot be solved by factoring quadratics. Answer is incorrect.

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3) \( 4(x + 3)(x + 1) = 0 \)

Set each factor to zero:
- \( x + 3 = 0 \Rightarrow x = -3 \)
- \( x + 1 = 0 \Rightarrow x = -1 \)

The constant 4 doesn’t affect roots.

✔ Solution: \( \{-3, -1\} \) — Correct

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4) \( (2x - 3)(x + 3) = 0 \)

Set each factor to zero:
- \( 2x - 3 = 0 \Rightarrow x = \frac{3}{2} \)
- \( x + 3 = 0 \Rightarrow x = -3 \)

✔ Solution: \( \left\{\frac{3}{2}, -3\right\} \) — Correct

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5) \( x^2 + 11x + 10 = -5 \)

First, bring all terms to one side:
\( x^2 + 11x + 10 + 5 = 0 \Rightarrow x^2 + 11x + 15 = 0 \)

Now factor:
Look for two numbers that multiply to 15 and add to 11:
→ 5 and 3? 5×3=15, 5+3=8 ✘
→ 15 and 1: 15+1=16 ✘
→ No integer factors → Not factorable over integers.

Try quadratic formula:
\( x = \frac{-11 \pm \sqrt{121 - 60}}{2} = \frac{-11 \pm \sqrt{61}}{2} \)

But the answer says \( \{2, 6\} \). Let’s test:
- \( x = 2 \): \( 4 + 22 + 10 = 36 \neq -5 \) ✘
- \( x = 6 \): \( 36 + 66 + 10 = 112 \neq -5 \) ✘

✘ Incorrect solution

Wait — perhaps the original equation was:
\( x^2 + 11x + 10 = 0 \)? Then roots would be:
\( (x+10)(x+1)=0 \Rightarrow x=-10,-1 \) — still not {2,6}

Or maybe:
\( x^2 - 8x + 12 = 0 \Rightarrow (x-6)(x-2)=0 \Rightarrow x=2,6 \)

So likely the equation was \( x^2 - 8x + 12 = 0 \), not what’s written.

✔ Conclusion: Given equation leads to non-integer roots; answer is incorrect.

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6) \( x^2 + 7x + 12 = 0 \)

Factor:
Find two numbers that multiply to 12 and add to 7: 3 and 4
→ \( (x+3)(x+4) = 0 \)
→ \( x = -3, -4 \)

Given answer: \( \{-6, -2\} \) — check:
- \( (-6)^2 + 7(-6) + 12 = 36 - 42 + 12 = 6 \neq 0 \) ✘

✘ Incorrect

Correct solution: \( \{-3, -4\} \)

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7) \( x^2 - 10x + 22 = -2 \)

Bring all terms to one side:
\( x^2 - 10x + 22 + 2 = 0 \Rightarrow x^2 - 10x + 24 = 0 \)

Factor:
Two numbers that multiply to 24 and add to -10: -6 and -4
→ \( (x - 6)(x - 4) = 0 \)
→ \( x = 6, 4 \)

Given answer: \( \{6, 4\} \) — matches ✔

✔ Correct

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8) \( x^2 + 3x + 12 = 6 \)

Bring to zero:
\( x^2 + 3x + 6 = 0 \)

Discriminant: \( 9 - 24 = -15 \) → no real solutions

But answer says \( \{3, -6\} \) — test:
- \( x = 3 \): \( 9 + 9 + 12 = 30 \neq 6 \) ✘
- \( x = -6 \): \( 36 - 18 + 12 = 30 \neq 6 \) ✘

✘ Incorrect

Wait — if equation was \( x^2 + 3x - 18 = 0 \), then:
\( (x+6)(x-3)=0 \Rightarrow x=-6,3 \)

So likely typo: equation should be \( x^2 + 3x - 18 = 0 \)

✔ Answer is correct only if equation is changed

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9) \( 6x^2 - 18x - 18 = 6 \)

Bring to zero:
\( 6x^2 - 18x - 24 = 0 \)

Divide entire equation by 6:
\( x^2 - 3x - 4 = 0 \)

Factor:
\( (x - 4)(x + 1) = 0 \) → \( x = 4, -1 \)

Given answer: \( \{4, -1\} \) — ✔ Correct

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10) \( 7x^2 - 14x = -7 \)

Bring to zero:
\( 7x^2 - 14x + 7 = 0 \)

Divide by 7:
\( x^2 - 2x + 1 = 0 \)

Factor:
\( (x - 1)^2 = 0 \) → \( x = 1 \)

Given answer: \( \{1\} \) — ✔ Correct

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✔ Summary of Corrections:

| Problem | Given Answer | Correct? | Notes |
|--------|--------------|---------|-------|
| 1 | {-1, 5} | ✔ Yes | Correct |
| 2 | {-4, -2} | ✘ No | Equation is linear, not quadratic |
| 3 | {-3, -1} | ✔ Yes | Correct |
| 4 | {3/2, -3} | ✔ Yes | Correct |
| 5 | {2, 6} | ✘ No | Equation likely mistyped |
| 6 | {-6, -2} | ✘ No | Should be {-3, -4} |
| 7 | {6, 4} | ✔ Yes | Correct |
| 8 | {3, -6} | ✘ No | Equation likely mistyped |
| 9 | {4, -1} | ✔ Yes | Correct |
| 10 | {1} | ✔ Yes | Correct |

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🔍 Final Thoughts:

Several problems have typographical errors in either the equation or the solution. Most likely, the original worksheet had mistakes.

To solve quadratic equations by factoring:
1. Write the equation in standard form: \( ax^2 + bx + c = 0 \)
2. Factor the left-hand side
3. Use the Zero Product Property: set each factor equal to zero
4. Solve for \( x \)

Let me know if you’d like me to rewrite the corrected version of the worksheet!