Expanding and factoring polynomials

Expanding and factoring are one skill seen from two directions. Expanding multiplies a product out into a sum of terms; factoring runs the same multiplication backward, turning a sum of terms into a product. A short set of patterns covers almost everything the SAT asks: multiplying two binomials term by term, squaring a sum into a trinomial with a doubled middle term, and recognizing a difference of squares. The work is quick, so the points leak in four predictable places: the middle term of a square dropped, a difference of squares mishandled, a sign or a term slipped during the multiply or the factor, and the right algebra reported as the wrong object, a root when a factor was asked or the reverse.

Expanding a product means multiplying each term in the first factor by each term in the second, then combining like terms. For two binomials that is four products, and the two cross products in the middle are where points go. Squaring a sum is the same move: write it as the binomial times itself, so the middle term comes out doubled. The single most common slip is squaring a sum term by term and losing that middle term entirely.

• Two binomials: multiply all four pairs. $(x+a)(x+b)=x^{2}+(a+b)x+ab$.
• A squared sum keeps a doubled middle term. $(x+a)^{2}=x^{2}+2ax+a^{2}$.
• A squared sum is never the sum of the squares. $(x+a)^{2}\neq x^{2}+a^{2}$.

Worked example. Expand $(x+4)^{2}$.

When an expression is one perfect square minus another, it always factors into conjugates: the same two square roots, once added and once subtracted. The middle terms cancel, which is why the product has no middle term to begin with. Two cautions live here. A sum of squares, $x^{2}+a^{2}$, does not factor over the real numbers; it is not a difference, so the rule does not apply. And a difference of squares is not a perfect square: $x^{2}-25$ is $(x+5)(x-5)$, not $(x-5)^{2}$, which would carry a middle term the original never had.

• Difference of squares: $a^{2}-b^{2}=(a+b)(a-b)$, with no middle term.
• A coefficient is fine: $4x^{2}-25=(2x+5)(2x-5)$.
• A sum of squares does not factor: $x^{2}+9$ stays as it is over the reals.

Worked example. Factor $x^{2}-25$.

To factor $x^{2}+bx+c$, find two numbers that multiply to $c$ and add to $b$. Their signs are the whole game. If $c$ is positive, the two numbers share a sign, the sign of $b$. If $c$ is negative, they have opposite signs, and the one larger in size carries the sign of $b$. Choosing the right pair of numbers but the wrong signs is the slip that flips the middle term and lands on a close, wrong answer.

• Multiply to the constant, add to the middle coefficient: $(x+p)(x+q)$ with $pq=c$ and $p+q=b$.
• Positive constant: both factors share the sign of $b$. $x^{2}-5x+6=(x-2)(x-3)$.
• Negative constant: opposite signs. $x^{2}+x-6=(x+3)(x-2)$.

Worked example. Factor $x^{2}-5x+6$.

Once the factoring is done, the question decides what to hand back, and the pieces look alike enough to swap by accident. A factor is a linear expression such as $x-3$. A root is the value that makes a factor zero, found by flipping the sign of its constant, so $x-3$ has root $3$ and $x+3$ has root $-3$. Solving an equation asks for roots; an item asking which expression is a factor wants the binomial itself. On a grid-in, watch whether the item wants a coefficient or a constant. Each is a real number from the same problem, but only one answers the question asked.

Worked example. Given $x^{2}+x-6=(x+3)(x-2)$, what are the solutions to $x^{2}+x-6=0$?

Multiply every pair and keep the middle term, factor a difference of squares into conjugates while leaving a sum of squares alone, hold every sign steady, then report the factor, root, or value the item actually names.

Four patterns recur on expanding and factoring. They are the same ones the diagnostic routes on.