Factoring Quadratic Equations - Free Printable

This image displays the answers to “Factorising Quadratic Equations – Sheet 10”. It does not contain a problem to solve — it is already solved and shows the correct factorisations and solutions for 16 quadratic equations.

However, I can explain how to solve each type of problem shown here, using the first few as examples. The goal is to factorise a quadratic equation (usually in the form \( ax^2 + bx + c = 0 \)) into two binomials, then use the zero product property to find the roots (solutions).

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🔍 General Method:

For a quadratic equation:
\[ ax^2 + bx + c = 0 \]

1. Factorise it into the form:
\[ (px + q)(rx + s) = 0 \]
2. Set each factor equal to zero:
\( px + q = 0 \) → solve for \( x \)
\( rx + s = 0 \) → solve for \( x \)
3. Write both solutions.

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## 📝 Step-by-Step Examples:

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✔ Problem 1: \( 2x^2 + 3x + 1 = 0 \)

Step 1: Factorise

We need two numbers that multiply to \( 2 \times 1 = 2 \), and add to 3 → those are 2 and 1.

Split middle term:
\( 2x^2 + 2x + x + 1 = 0 \)

Group:
\( (2x^2 + 2x) + (x + 1) = 0 \)
→ \( 2x(x + 1) + 1(x + 1) = 0 \)
→ \( (2x + 1)(x + 1) = 0 \)

Step 2: Solve

Set each factor = 0:

- \( 2x + 1 = 0 \) → \( x = -\frac{1}{2} \)
- \( x + 1 = 0 \) → \( x = -1 \)

✔ Final Answer: \( x = -\frac{1}{2} \) or \( -1 \)

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✔ Problem 2: \( 2y^2 + 7y + 3 = 0 \)

Factorise:

Multiply \( 2 \times 3 = 6 \). Need two numbers that multiply to 6 and add to 7 → 6 and 1.

Split:
\( 2y^2 + 6y + y + 3 = 0 \)
→ \( 2y(y + 3) + 1(y + 3) = 0 \)
→ \( (2y + 1)(y + 3) = 0 \)

Solve:

- \( 2y + 1 = 0 \) → \( y = -\frac{1}{2} \)
- \( y + 3 = 0 \) → \( y = -3 \)

✔ Final Answer: \( y = -\frac{1}{2} \) or \( -3 \)

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✔ Problem 3: \( 2z^2 - 8z + 8 = 0 \)

First, simplify by dividing all terms by 2:

→ \( z^2 - 4z + 4 = 0 \)

This is a perfect square:

→ \( (z - 2)^2 = 0 \) → so \( z = 2 \) (repeated root)

But the answer sheet shows: \( (2z - 4)(z - 2) = 0 \)

Let’s check that:

\( (2z - 4)(z - 2) = 2z(z - 2) -4(z - 2) = 2z^2 - 4z -4z + 8 = 2z^2 - 8z + 8 \) ✓

Now solve:

- \( 2z - 4 = 0 \) → \( z = 2 \)
- \( z - 2 = 0 \) → \( z = 2 \)

So only one unique solution: \( z = 2 \)

✔ Final Answer: \( z = 2 \)

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✔ Problem 4: \( 2a^2 - 3a - 5 = 0 \)

Multiply \( 2 \times -5 = -10 \). Need two numbers that multiply to -10 and add to -3 → -5 and 2

Split:
\( 2a^2 - 5a + 2a - 5 = 0 \)
→ \( a(2a - 5) + 1(2a - 5) = 0 \)
→ \( (2a - 5)(a + 1) = 0 \)

Solve:

- \( 2a - 5 = 0 \) → \( a = \frac{5}{2} \)
- \( a + 1 = 0 \) → \( a = -1 \)

✔ Final Answer: \( a = \frac{5}{2} \) or \( -1 \)

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⚠️ Common Mistake to Avoid:

Sometimes students forget to set each factor equal to zero or mis-solve the linear equations.

Example: In problem 9:
\( (2f + 2)(f - 4) = 0 \)

Some might say “f = 4” and forget the other factor.

But:
- \( 2f + 2 = 0 \) → \( f = -1 \)
- \( f - 4 = 0 \) → \( f = 4 \)

✔ So solutions are \( f = -1 \) or \( 4 \)

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## 🧩 Pattern Recognition:

All problems follow the same structure:

- Leading coefficient is often 2.
- You factor by splitting the middle term.
- Then solve each linear factor.

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## ✔ Summary:

The image you uploaded is already the answer key. If you were meant to solve these problems yourself, you would:

1. Factorise each quadratic.
2. Set each factor = 0.
3. Solve for the variable.
4. Write both solutions (unless repeated root).

If you’d like me to walk through any specific problem from the list in more detail, just ask!

Otherwise, this sheet is correctly solved and ready for checking your own work. ✔️