Factoring Quadratics: Practice

To factor $x^2 + 5x + 6$ you need two numbers that add to $5$ and multiply to $6$. Test the positive pairs that multiply to $6$.

$2 + 3 = 5 \quad \text{and} \quad 2 \cdot 3 = 6$

So the pair is $2$ and $3$. The pair $1$ and $6$ multiplies to $6$ but adds to $7$, and $-2$ and $-3$ add to $-5$.

Find two numbers with sum $5$ and product $6$. Both are positive, and $2$ and $3$ work.

$x^2 + 5x + 6 = (x + 2)(x + 3)$

Expanding back gives $x^2 + 3x + 2x + 6 = x^2 + 5x + 6$, so the factoring is correct.

Two positive numbers with sum $6$ and product $8$ are $2$ and $4$.

$x^2 + 6x + 8 = (x + 2)(x + 4)$

The pair $1$ and $8$ has product $8$ but sum $9$, so only $2$ and $4$ work.

By the zero-product property, set each factor to $0$.

$x + 1 = 0 \quad \text{or} \quad x + 4 = 0$

So $x = -1$ or $x = -4$. Each root is the opposite sign of the number shown in its factor.

Set each factor equal to $0$.

$x - 2 = 0 \quad \text{or} \quad x - 3 = 0$

So $x = 2$ or $x = 3$.

The product $3$ is positive and the sum $-4$ is negative, so both numbers are negative, namely $-1$ and $-3$.

$x^2 - 4x + 3 = (x - 1)(x - 3)$

Expanding gives $x^2 - 3x - x + 3 = x^2 - 4x + 3$.

Both terms contain $x$, so $x$ is the common factor. Pull it out front.

$x^2 + 3x = x(x + 3)$

Check by expanding: $x(x + 3) = x^2 + 3x$, the original.

Two positive numbers with sum $10$ and product $21$ are $3$ and $7$.

$x^2 + 10x + 21 = (x + 3)(x + 7)$

Expanding $(x + 3)(x + 7)$ gives $x^2 + 7x + 3x + 21 = x^2 + 10x + 21$, so only this one matches.

The expression is already a product of the factors $x$ and $x - 6$. Set each to $0$.

$x = 0 \quad \text{or} \quad x - 6 = 0$

So $x = 0$ or $x = 6$. Do not lose the root $x = 0$ from the first factor.

Two numbers with sum $2$ and product $1$ are $1$ and $1$.

$x^2 + 2x + 1 = (x + 1)(x + 1)$

Expanding gives $x^2 + x + x + 1 = x^2 + 2x + 1$.

The product $12$ is positive and the sum $-7$ is negative, so both numbers are negative.

$-3 + (-4) = -7 \quad \text{and} \quad (-3)(-4) = 12$

So the pair is $-3$ and $-4$.

Set each factor equal to $0$.

$x + 5 = 0 \quad \text{or} \quad x - 2 = 0$

So $x = -5$ or $x = 2$. Each root has the opposite sign of the number in its factor.