Factoring Quadratics: Practice

The product $-15$ is negative, so the signs are opposite, and the sum $-2$ is negative, so the larger number is negative. They are $-5$ and $3$.

$x^2 - 2x - 15 = (x - 5)(x + 3)$

Only the sign on the larger number changes from the $+2$ version of this trinomial.

The product $-10$ is negative and the sum is $3$, so the numbers are $5$ and $-2$.

$x^2 + 3x - 10 = (x + 5)(x - 2) = 0$

Setting each factor to $0$ gives $x = -5$ or $x = 2$.

Every term is even, so pull out the common factor $2$ first.

$2x^2 + 14x + 20 = 2(x^2 + 7x + 10) = 2(x + 2)(x + 5)$

Inside, two numbers with sum $7$ and product $10$ are $2$ and $5$. Dropping the $2$ leaves the factoring incomplete.

There is no constant term, so pull out the shared $x$ rather than dividing by it.

$x^2 - 5x = x(x - 5) = 0$

Setting each factor to $0$ gives $x = 0$ or $x = 5$. Dividing by $x$ would throw away the root $x = 0$.

The product $-12$ is negative and the sum is $-1$, so the numbers are $3$ and $-4$.

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

Expanding gives $x^2 - 4x + 3x - 12 = x^2 - x - 12$, so this is the only match.

Two numbers with sum $7$ and product $12$ are $3$ and $4$.

$x^2 + 7x + 12 = (x + 3)(x + 4) = 0$

Setting each factor to $0$ gives $x = -3$ or $x = -4$.

The surest check is to expand the factors and see whether they rebuild the original trinomial.

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

The result matches, so the factoring is correct. A single substitution only shows the two forms agree at that one value of $x$, while expanding proves they are equal for every $x$.

The product $-20$ is negative and the sum is $-1$, so the numbers are $-5$ and $4$.

$x^2 - x - 20 = (x - 5)(x + 4) = 0$

Setting each factor to $0$ gives $x = 5$ or $x = -4$.

For $x^2 + x + 1$ you would need two integers with product $1$ and sum $1$, but the only integer pairs multiplying to $1$ are $1, 1$ (sum $2$) and $-1, -1$ (sum $-2$).

$\text{no integer pair has product } 1 \text{ and sum } 1$

So it is irreducible over the integers. The other three factor as $(x+2)(x+3)$, $(x-2)(x-3)$, and $(x+1)(x+6)$.

The greatest common factor of $3x^2$ and $12x$ is $3x$. Pull it out.

$3x^2 - 12x = 3x(x - 4)$

The forms $3(x^2 - 4x)$ and $x(3x - 12)$ are only partly factored, since each still hides a shared factor inside.

The product $-12$ is negative and the sum is $+4$, so the larger number is positive. The numbers are $6$ and $-2$.

$x^2 + 4x - 12 = (x + 6)(x - 2)$

Expanding gives $x^2 - 2x + 6x - 12 = x^2 + 4x - 12$.

Move every term to one side: $x^2 - 5x - 24 = 0$. Two numbers with product $-24$ and sum $-5$ are $3$ and $-8$.

$x^2 - 5x - 24 = (x + 3)(x - 8) = 0$

Setting each factor to $0$ gives $x = -3$ or $x = 8$.