Squares of Binomials: Free Response

Use $(a+b)^2=a^2+2ab+b^2$ with $a=x$ and $b=13$. Square the first term, square the last term, and put twice their product in the middle.

$(x+13)^2 = x^2 + 2\cdot x\cdot 13 + 13^2 = x^2 + 26x + 169.$

Use $(a-b)^2=a^2-2ab+b^2$ with $a=x$ and $b=13$. The first and last terms are the same squares as before; only the middle term's sign changes.

$(x-13)^2 = x^2 - 2\cdot x\cdot 13 + 13^2 = x^2 - 26x + 169.$

Line the two results up:

$x^2 + 26x + 169 \qquad \text{and} \qquad x^2 - 26x + 169.$

The first term $x^2$ and the last term $169$ match exactly in both. Only the middle term differs, and the reason traces back to where it comes from: in $(x+13)(x+13)$ the two cross-products are $x\cdot 13=13x$ and $13\cdot x=13x$, both positive, and they combine to $+26x$. In $(x-13)(x-13)$ the cross-products are $x\cdot(-13)=-13x$ and $(-13)\cdot x=-13x$, both negative, combining to $-26x$. The first and last terms never involve a sign choice, since $13^2$ and $(-13)^2$ are both positive, but the cross-products carry the sign of $b$ straight through.

Use $(a+b)^2=a^2+2ab+b^2$ with $a=3x$ and $b=7$. Square the WHOLE first term, including its coefficient.

$(3x+7)^2 = (3x)^2 + 2\cdot(3x)\cdot 7 + 7^2 = 9x^2 + 42x + 49.$

Apply the squared-difference pattern to the two full terms $4a$ and $3b$ (these letters are the TERMS being squared, not the $a,b$ of the pattern itself).

$(4a-3b)^2 = (4a)^2 - 2\cdot(4a)\cdot(3b) + (3b)^2 = 16a^2 - 24ab + 9b^2.$

Squaring $kx$ means multiplying it by itself:

$(kx)^2 = (kx)(kx) = k\cdot x\cdot k\cdot x.$

Multiplication can be reordered and regrouped freely, so gather the two $k$'s and the two $x$'s together:

$k\cdot x\cdot k\cdot x = (k\cdot k)\cdot(x\cdot x) = k^2x^2.$

That argument never used a specific value of $k$, so it holds for every coefficient there is, not just $3$ or $4$ from parts A and B. Writing the first term as $kx^2$ instead treats only the $x$ as squared and leaves the coefficient at its original power, $k^1$ instead of $k^2$: it is missing exactly one factor of $k$. For $k=1$ the two versions happen to agree, since $k^2=k$ there, but for every other value they genuinely differ, which is why the shortcut fails in general and not just occasionally.

The whole patio is a square of side $x+9$, so its area is $(x+9)^2$. Apply $(a+b)^2=a^2+2ab+b^2$ with $a=x$, $b=9$:

$(x+9)^2 = x^2 + 2\cdot x\cdot 9 + 9^2 = x^2 + 18x + 81.$

Each tile contributes one term:

$x\cdot x = x^2, \qquad x\cdot 9 = 9x \text{ (twice)}, \qquad 9\cdot 9 = 81.$

Adding all four,

$x^2 + 9x + 9x + 81 = x^2 + 18x + 81,$

since the two $9x$ strips combine into $18x$. That is exactly the trinomial from part A, matched term for term: the lone tile gives the first square, the corner tile gives the last square, and the TWO strips together give the middle term.

With only one strip, the three tiles would add to

$x^2 + 9x + 81,$

which is short of the correct total $x^2+18x+81$ by exactly one $9x$ strip's worth of area.

This is not a quirk of this one patio. In the identity $(a+b)^2=a^2+2ab+b^2$, the middle term $2ab$ exists because a square of side $a+b$ genuinely splits into TWO equal $a$-by-$b$ rectangles, not one: one running along the top of the corner square and one running along its side. They are two separate physical regions with the same area, and both belong to the total. Counting only one of them, as the three-tile patio does, always leaves the area short by exactly one copy of $ab$, no matter what $a$ and $b$ are.

A claim quantified over ALL real numbers is destroyed by a single pair on which it fails, so pick one and evaluate both sides honestly.

Take $a=6$ and $b=2$. The left side of the claim:

$(a-b)^2 = (6-2)^2 = 4^2 = 16.$

The right side of the claim:

$a^2 - b^2 = 6^2 - 2^2 = 36 - 4 = 32.$

The two sides give $16$ and $32$, which are not equal, so this pair refutes the claim.

Apply the squared-difference pattern:

$(a-b)^2 = a^2 - 2ab + b^2.$

Compare this correct expansion against the claim, $a^2-b^2$. Two things are wrong with the claim: it has no middle term at all, when the true expansion has $-2ab$, and its last term is written as $-b^2$, when $b^2$ is a square and squares are never negative, so the true last term is $+b^2$.

Set the two sides of the claim equal and solve for when they genuinely agree:

$a^2 - 2ab + b^2 = a^2 - b^2.$

Subtract $a^2$ from both sides and move everything to one side:

$-2ab + b^2 = -b^2 \quad\Longrightarrow\quad 2b^2 - 2ab = 0 \quad\Longrightarrow\quad 2b(b-a)=0.$

A product is zero only when a factor is, so $b=0$ or $b=a$. Those are the only two families of pairs where the claim happens to be true: when $b$ is zero, or when $a$ and $b$ are equal.

This does not rescue the general claim. "For all real numbers $a$ and $b$" asserts the equality on EVERY pair, and part A already produced one pair, $a=6$, $b=2$ (neither $b=0$ nor $a=b$), where it fails. One counterexample is enough to make a universal claim false; finding pairs where it happens to hold afterward does not undo that, it only maps out the narrow region where the false general claim accidentally gives the right answer.

Every coefficient is divisible by $2$, so pull it out first.

$2x^2+28x+98 = 2\left(x^2+14x+49\right).$

Inside the parentheses, test the three-part rule: $\sqrt{x^2}=x$ and $\sqrt{49}=7$, so $A=x$, $B=7$, and $2AB=2\cdot x\cdot 7=14x$, which matches the middle term with a positive sign. So $x^2+14x+49=(x+7)^2$, and the complete factorization is

$2x^2+28x+98 = 2(x+7)^2.$

From part A, $2x^2+28x+98 = 2(x+7)^2$, so the equation becomes

$2(x+7)^2 = 0.$

Dividing by the nonzero constant $2$ leaves $(x+7)^2=0$, that is, $(x+7)(x+7)=0$. By the zero-product property, $x+7=0$ or $x+7=0$: the SAME equation twice, giving

$x=-7.$

There is only one distinct solution, even though the squared factor formally produces two factor equations; they are not different equations.

Start from $k(A+B)^2=0$ with $k\neq 0$. Dividing both sides by $k$ (legal since $k\neq 0$) leaves

$(A+B)^2 = 0, \qquad\text{that is,}\qquad (A+B)(A+B)=0.$

The zero-product property says a product is zero exactly when at least one factor is, so it sets EACH of the two factors to zero:

$A+B=0 \qquad \text{or} \qquad A+B=0.$

Both factors are the same expression, so these are not two different conditions, they are the same linear equation written twice. A single linear equation in one unknown pins down exactly one value. So no matter what $A$, $B$, or nonzero $k$ are, a perfect square set to zero can never produce two genuinely different solutions: the squaring itself is what collapses two factor slots into one repeated condition.