Quadratic Expressions

1. We need two numbers that multiply to $12$ and add to $7$. Those numbers are $3$ and $4$.

$x^2 + 7x + 12 = (x + 3)(x + 4)$

Check: $(x + 3)(x + 4) = x^2 + 4x + 3x + 12 = x^2 + 7x + 12$ ✓

2. First, factor out the greatest common factor of $2$:

$2x^2 - 10x - 12 = 2(x^2 - 5x - 6)$

Now factor the trinomial. We need two numbers that multiply to $-6$ and add to $-5$. Those numbers are $-6$ and $1$.

$2(x^2 - 5x - 6) = 2(x - 6)(x + 1)$

Check: $2(x - 6)(x + 1) = 2(x^2 + x - 6x - 6) = 2(x^2 - 5x - 6) = 2x^2 - 10x - 12$ ✓

3. We need two numbers that multiply to $20$ and add to $-9$. Those numbers are $-4$ and $-5$.

$f(x) = x^2 - 9x + 20 = (x - 4)(x - 5)$

The zeros occur when $f(x) = 0$:

$(x - 4)(x - 5) = 0$ $x - 4 = 0 \quad \text{or} \quad x - 5 = 0$ $x = 4 \quad \text{or} \quad x = 5$

Check: $f(4) = 16 - 36 + 20 = 0$ ✓ and $f(5) = 25 - 45 + 20 = 0$ ✓

4. The expression $4x^2 - 25$ is a difference of two squares. We recognize that $4x^2 = (2x)^2$ and $25 = 5^2$, so we apply the identity $a^2 - b^2 = (a + b)(a - b)$:

$4x^2 - 25 = (2x)^2 - 5^2 = (2x + 5)(2x - 5)$

Check: $(2x + 5)(2x - 5) = 4x^2 - 10x + 10x - 25 = 4x^2 - 25$ ✓

5. a) We need two numbers that multiply to $-28$ and add to $3$. Those numbers are $7$ and $-4$.

$A(x) = x^2 + 3x - 28 = (x + 7)(x - 4)$

Check: $(x + 7)(x - 4) = x^2 - 4x + 7x - 28 = x^2 + 3x - 28$ ✓

b) Since $x$ represents a length, $x > 0$. The factors represent dimensions, so both must be positive:

• $x + 7 > 0$ is true for all $x > 0$
• $x - 4 > 0$ requires $x > 4$

Therefore $x > 4$ for the area to be positive and both dimensions to represent valid lengths.

When $x = 7$:

• One side: $x + 7 = 7 + 7 = 14$ feet
• Other side: $x - 4 = 7 - 4 = 3$ feet
• Area: $A(7) = 14 \times 3 = 42$ square feet

Check: $A(7) = 49 + 21 - 28 = 42$ ✓