Solving Quadratics by Factoring Worksheet - Classful - Free Printable

Let's solve each of the quadratic equations by factoring, as instructed. We'll go through each one step-by-step.

---

Question 1:

Equation: $ x^2 - 5x - 14 = 0 $

We need two numbers that multiply to $-14$ and add to $-5$.
Those numbers are $-7$ and $2$:
$$
x^2 - 7x + 2x - 14 = 0 \\
x(x - 7) + 2(x - 7) = 0 \\
(x + 2)(x - 7) = 0
$$

Set each factor equal to zero:
$ x + 2 = 0 \Rightarrow x = -2 $
$ x - 7 = 0 \Rightarrow x = 7 $

✔ Solutions: $ x = -2, 7 $

---

Question 2:

Equation: $ 2x^2 - 3x - 5 = 0 $

We use factoring by grouping or AC method.
AC = $ 2 \times (-5) = -10 $. Find two numbers that multiply to $-10$ and add to $-3$:
They are $-5$ and $2$

Rewrite middle term:
$$
2x^2 - 5x + 2x - 5 = 0 \\
x(2x - 5) + 1(2x - 5) = 0 \\
(2x - 5)(x + 1) = 0
$$

Set each factor to zero:
$ 2x - 5 = 0 \Rightarrow x = \frac{5}{2} $
$ x + 1 = 0 \Rightarrow x = -1 $

✔ Solutions: $ x = -1, \frac{5}{2} $

---

Question 3:

Equation: $ 4x^2 + 11x - 20 = 0 $

AC = $ 4 \times (-20) = -80 $. Need two numbers that multiply to $-80$, add to $11$:
Try: $16$ and $-5$: $16 \times (-5) = -80$, $16 + (-5) = 11$ ✔

Rewrite:
$$
4x^2 + 16x - 5x - 20 = 0 \\
4x(x + 4) -5(x + 4) = 0 \\
(4x - 5)(x + 4) = 0
$$

Solve:
$ 4x - 5 = 0 \Rightarrow x = \frac{5}{4} $
$ x + 4 = 0 \Rightarrow x = -4 $

✔ Solutions: $ x = -4, \frac{5}{4} $

---

Question 4:

Equation: $ 2n^2 - 5n + 3 = 0 $

AC = $ 2 \times 3 = 6 $. Need two numbers that multiply to $6$, add to $-5$:
$-3$ and $-2$: $(-3)(-2)=6$, $-3 + (-2) = -5$ ✔

Rewrite:
$$
2n^2 - 3n - 2n + 3 = 0 \\
n(2n - 3) -1(2n - 3) = 0 \\
(n - 1)(2n - 3) = 0
$$

Solve:
$ n - 1 = 0 \Rightarrow n = 1 $
$ 2n - 3 = 0 \Rightarrow n = \frac{3}{2} $

✔ Solutions: $ n = 1, \frac{3}{2} $

---

Question 5:

Equation: $ x^2 - 5x = 2 $

First, move all terms to one side:
$ x^2 - 5x - 2 = 0 $

Now try to factor:
Look for two numbers that multiply to $-2$ and add to $-5$:
No such integers exist. Try discriminant:
$ D = (-5)^2 - 4(1)(-2) = 25 + 8 = 33 $ → Not a perfect square ⇒ Not factorable over integers

⚠️ Cannot be factored easily. Use quadratic formula instead.
But since the instruction says "solve by factoring", this one might not have integer solutions.

Wait — double-check: maybe I made a mistake?

Try: factors of $-2$: $1, -2$; $-1, 2$
Can't get sum $-5$. So indeed, not factorable with integers.

✘ This equation cannot be solved by factoring using integers.
So perhaps there’s a typo, or it's meant to be solved differently.

But let's assume we're only solving factorable ones. Maybe skip or note.

Alternatively, check if it was supposed to be $=0$? But no, it's $=2$. So unless it's acceptable to complete the square or use quadratic formula, we can’t factor it nicely.

✔ Conclusion: This one is not factorable over integers.

---

Question 6:

Equation: $ x^2 = x + 3 $

Bring all terms to one side:
$ x^2 - x - 3 = 0 $

Try to factor:
Need two numbers multiplying to $-3$, adding to $-1$:
$-3, 1$: $-3+1=-2$ ✘
$3, -1$: $3 + (-1) = 2$ ✘
No pair works.

Discriminant: $ (-1)^2 - 4(1)(-3) = 1 + 12 = 13 $ → Not a perfect square

✘ Not factorable over integers

---

Question 7:

Equation: $ x^2 - 4x = -4 $

Bring to standard form:
$ x^2 - 4x + 4 = 0 $

Factor:
This is a perfect square trinomial:
$ (x - 2)^2 = 0 $

So:
$ x - 2 = 0 \Rightarrow x = 2 $

✔ Solution: $ x = 2 $ (double root)

---

Question 8:

Equation: $ x^2 - 3x - 9 = 3x $

Move all terms to left:
$ x^2 - 3x - 9 - 3x = 0 \Rightarrow x^2 - 6x - 9 = 0 $

Check for factoring:
Need two numbers multiplying to $-9$, adding to $-6$:
Possible pairs: $-9,1$ → sum $-8$; $-3,3$ → sum $0$; $-1,9$ → sum $8$
None give $-6$

Discriminant: $ (-6)^2 - 4(1)(-9) = 36 + 36 = 72 $ → Not a perfect square

✘ Not factorable over integers

---

Question 9:

Equation: $ n^2 - 7n - 13 = -10 $

Bring to standard form:
$ n^2 - 7n - 13 + 10 = 0 \Rightarrow n^2 - 7n - 3 = 0 $

Check for factoring:
Need numbers multiplying to $-3$, adding to $-7$:
$-3,1$ → sum $-2$; $3,-1$ → sum $2$ → none work

Discriminant: $ 49 + 12 = 61 $ → Not perfect square

✘ Not factorable over integers

---

Question 10:

Equation: $ 3x^2 + 2x = 5 $

Bring to standard form:
$ 3x^2 + 2x - 5 = 0 $

AC = $ 3 \times (-5) = -15 $. Need two numbers that multiply to $-15$, add to $2$:
Try: $5$ and $-3$: $5 + (-3) = 2$ ✔

Rewrite:
$$
3x^2 + 5x - 3x - 5 = 0 \\
x(3x + 5) -1(3x + 5) = 0 \\
(x - 1)(3x + 5) = 0
$$

Solve:
$ x - 1 = 0 \Rightarrow x = 1 $
$ 3x + 5 = 0 \Rightarrow x = -\frac{5}{3} $

✔ Solutions: $ x = 1, -\frac{5}{3} $

---

Question 11:

Equation: $ 9x^2 = 7x + 4 $

Bring to standard form:
$ 9x^2 - 7x - 4 = 0 $

AC = $ 9 \times (-4) = -36 $. Need two numbers that multiply to $-36$, add to $-7$:
Try: $-9$ and $4$: $-9 + 4 = -5$ ✘
$-12$ and $3$: $-12 + 3 = -9$ ✘
$-4$ and $9$: $-4 + 9 = 5$ ✘
$-3$ and $12$: $-3 + 12 = 9$ ✘
$-18$ and $2$: $-18 + 2 = -16$ ✘

Wait: $-9$ and $4$: no. What about $-9$ and $4$? No.

Try: $-9$ and $4$ → product $-36$, sum $-5$
Try $-12$ and $3$: $-9$ → no

Try $-4$ and $9$: sum $5$
Try $-3$ and $12$: sum $9$

Wait: what about $-9$ and $4$? No.

Try $-18$ and $2$: sum $-16$

Try $-6$ and $6$: sum $0$

Wait: $-9$ and $4$: sum $-5$
How about $-12$ and $3$: sum $-9$

Wait: $-1$ and $36$: sum $35$
$1$ and $-36$: sum $-35$

Hmm. Try: $-9$ and $4$ → no

Wait: $-12$ and $3$ → no

Wait: $-4$ and $9$: sum $5$

Wait: $-3$ and $12$: sum $9$

Try $-18$ and $2$: sum $-16$

Wait — how about $-9$ and $4$? Still no.

Wait: $-9$ and $4$ → sum $-5$

What about $-12$ and $3$? Sum $-9$

Try $-6$ and $6$: sum $0$

Wait: try $-9$ and $4$ — no.

Wait: actually, let's try factoring directly:

Try: $ (9x + a)(x + b) = 9x^2 + (a + 9b)x + ab $

We want $ ab = -4 $, $ a + 9b = -7 $

Try $ a = 1, b = -4 $: $ a + 9b = 1 - 36 = -35 $ ✘
$ a = -1, b = 4 $: $ -1 + 36 = 35 $ ✘
$ a = 2, b = -2 $: $ 2 - 18 = -16 $ ✘
$ a = -2, b = 2 $: $ -2 + 18 = 16 $ ✘
$ a = 4, b = -1 $: $ 4 - 9 = -5 $ ✘
$ a = -4, b = 1 $: $ -4 + 9 = 5 $ ✘
$ a = 8, b = -0.5 $? Not integer.

Wait: $ a = -9, b = 1 $: $ ab = -9 $, no

Wait: $ a = -12, b = 1 $: $ ab = -12 $

Wait: maybe not factorable.

Try discriminant:
$ D = (-7)^2 - 4(9)(-4) = 49 + 144 = 193 $ → Not a perfect square

✘ Not factorable over integers

---

Question 12:

Equation: $ 4x^2 - 1 = -8x $

Bring to standard form:
$ 4x^2 + 8x - 1 = 0 $

Check factoring:
AC = $ 4 \times (-1) = -4 $. Need two numbers that multiply to $-4$, add to $8$:
Try: $4$ and $-1$: sum $3$
$-4$ and $1$: sum $-3$
$2$ and $-2$: sum $0$
No pair gives $8$

Discriminant: $ 64 - 4(4)(-1) = 64 + 16 = 80 $ → Not perfect square

✘ Not factorable over integers

---

## ✔ Summary of Solvable Ones (by factoring):

| Question | Equation | Factored Form | Solutions |
|--------|---------|----------------|-----------|
| 1 | $ x^2 - 5x - 14 = 0 $ | $ (x + 2)(x - 7) = 0 $ | $ x = -2, 7 $ |
| 2 | $ 2x^2 - 3x - 5 = 0 $ | $ (2x - 5)(x + 1) = 0 $ | $ x = -1, \frac{5}{2} $ |
| 3 | $ 4x^2 + 11x - 20 = 0 $ | $ (4x - 5)(x + 4) = 0 $ | $ x = -4, \frac{5}{4} $ |
| 4 | $ 2n^2 - 5n + 3 = 0 $ | $ (n - 1)(2n - 3) = 0 $ | $ n = 1, \frac{3}{2} $ |
| 7 | $ x^2 - 4x + 4 = 0 $ | $ (x - 2)^2 = 0 $ | $ x = 2 $ |
| 10 | $ 3x^2 + 2x - 5 = 0 $ | $ (x - 1)(3x + 5) = 0 $ | $ x = 1, -\frac{5}{3} $ |

---

## ✘ Unfactorable Over Integers:
- Q5: $ x^2 - 5x - 2 = 0 $
- Q6: $ x^2 - x - 3 = 0 $
- Q8: $ x^2 - 6x - 9 = 0 $
- Q9: $ n^2 - 7n - 3 = 0 $
- Q11: $ 9x^2 - 7x - 4 = 0 $
- Q12: $ 4x^2 + 8x - 1 = 0 $

These would require the quadratic formula or completing the square.

---

✔ Final Answer (Only Factorable Ones):

1. $ x = -2, 7 $
2. $ x = -1, \frac{5}{2} $
3. $ x = -4, \frac{5}{4} $
4. $ n = 1, \frac{3}{2} $
7. $ x = 2 $
10. $ x = 1, -\frac{5}{3} $

The rest cannot be factored using integers.

Let me know if you'd like the non-factorable ones solved using the quadratic formula!