Factoring Quadratics: Additional Practice (Set 2)

Each root $r$ comes from a factor $x - r$, so the root $-6$ gives $x + 6$ and the root $4$ gives $x - 4$.

$(x + 6)(x - 4) = 0$

Check: $x = -6$ makes the first factor $0$ and $x = 4$ makes the second $0$.

Every term is a multiple of $5$, so pull out the $5$ first.

$5x^2 + 25x + 30 = 5(x^2 + 5x + 6) = 5(x + 2)(x + 3)$

Inside, two numbers with sum $5$ and product $6$ are $2$ and $3$. Dropping the $5$ leaves the factoring incomplete.

Two positive numbers with sum $13$ and product $40$ are $5$ and $8$.

$x^2 + 13x + 40 = (x + 5)(x + 8)$

The pair $4$ and $10$ multiplies to $40$ but adds to $14$, so only $5$ and $8$ work.

The product $35$ is positive and the sum $-12$ is negative, so both numbers are negative, namely $-5$ and $-7$.

$x^2 - 12x + 35 = (x - 5)(x - 7)$

The pair $-1$ and $-35$ multiplies to $35$ but adds to $-36$, so it fails.

The product $-24$ is negative and the sum $-5$ is negative, so the larger number is negative. The numbers are $3$ and $-8$.

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

Expanding gives $x^2 - 8x + 3x - 24 = x^2 - 5x - 24$.

For $x^2 + 6x + 7$ you need two integers with product $7$ and sum $6$, but the integer pairs multiplying to $7$ are $1, 7$ (sum $8$) and $-1, -7$ (sum $-8$).

$\text{no integer pair has product } 7 \text{ and sum } 6$

So it is irreducible. The others factor as $(x+2)(x+4)$, $(x+3)^2$, and $(x+1)(x+5)$.

Two numbers with sum $10$ and product $25$ are $5$ and $5$.

$x^2 + 10x + 25 = (x + 5)(x + 5)$

Expanding gives $x^2 + 5x + 5x + 25 = x^2 + 10x + 25$.

Two positive numbers with sum $12$ and product $27$ are $3$ and $9$.

$x^2 + 12x + 27 = (x + 3)(x + 9)$

The pair $1$ and $27$ multiplies to $27$ but adds to $28$, so only $3$ and $9$ work.

Two numbers with sum $-8$ and product $16$ are $-4$ and $-4$.

$x^2 - 8x + 16 = (x - 4)(x - 4)$

Expanding gives $x^2 - 4x - 4x + 16 = x^2 - 8x + 16$.

Each root $r$ comes from a factor $x - r$, so the root $3$ gives $x - 3$ and the root $-8$ gives $x + 8$.

$(x - 3)(x + 8) = 0$

Check: $x = 3$ makes the first factor $0$ and $x = -8$ makes the second $0$.

The product $-30$ is negative and the sum $-7$ is negative, so the larger number is negative. The numbers are $3$ and $-10$.

$x^2 - 7x - 30 = (x + 3)(x - 10)$

Expanding gives $x^2 - 10x + 3x - 30 = x^2 - 7x - 30$.

Multiply every term of the first factor by every term of the second, then combine like terms.

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

The result matches the original trinomial, so the factoring is correct.