Difference of Squares: Free Response

Run all three conditions on each expression: exactly two terms, both perfect squares, and a minus sign between them.

For $16x^2-9$: two terms, $16x^2=(4x)^2$ and $9=3^2$ are both perfect squares, and they are subtracted. All three conditions hold, so it factors:

$16x^2-9=(4x)^2-3^2=(4x+3)(4x-3).$

For $x^2+121$: both terms are perfect squares and there are two of them, but they are ADDED, so the sign condition fails. This is a sum of squares and does not factor over the real numbers.

For $x^2-60$: there are two terms and the sign is a minus, but $60$ is not a perfect square, since no whole number squares to $60$. The perfect-square condition fails.

Take the square root of each whole term, coefficient and variable together.

For $49x^2-16y^2$, $\sqrt{49x^2}=7x$ and $\sqrt{16y^2}=4y$:

$49x^2-16y^2=(7x)^2-(4y)^2=(7x+4y)(7x-4y).$

For $100-9x^2$, $\sqrt{100}=10$ and $\sqrt{9x^2}=3x$. The number comes first this time, so the order has to carry through:

$100-9x^2=10^2-(3x)^2=(10+3x)(10-3x).$

The three-part test has three separate conditions, and part A shows that satisfying any two is not enough on its own.

$x^2+121$ satisfies the term-count condition (two terms) and the perfect-square condition (both terms are squares), and still fails, because the sign is a plus, not a minus. $x^2-60$ satisfies the term-count condition and the sign condition, and still fails, because $60$ is not itself a perfect square.

Only the full shape

$A^2-B^2=(A+B)(A-B)$

guarantees a factorization, so each of the three checks exists to confirm the expression truly has that shape before the identity gets applied. Dropping any one of the three checks leaves an expression that can look like a difference of squares without actually being one.

Factor the left side first.

$16x^2-81=(4x)^2-9^2=(4x+9)(4x-9).$

Set the product equal to zero and apply the zero-product property: a product is zero only when one of its factors is zero.

$(4x+9)(4x-9)=0 \implies 4x+9=0 \ \text{or} \ 4x-9=0.$

Solving each linear equation gives $x=-\tfrac{9}{4}$ or $x=\tfrac{9}{4}$.

Find the number exactly halfway between $68$ and $72$: that is $70$, with $68=70-2$ and $72=70+2$.

$68\times72=(70-2)(70+2)=70^2-2^2.$

Evaluate the square and subtract:

$70^2-2^2=4900-4=4896.$

In part A, the identity is a FACTORING step: writing $16x^2-81$ as $(4x+9)(4x-9)$ turns a single quadratic equation into two linear ones that the zero-product property can solve separately.

In part B, the identity is an ARITHMETIC shortcut: writing $68\times72$ as $(70-2)(70+2)$ replaces a two-digit multiplication with one square and one subtraction.

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

Both uses depend on the same structural fact: any pair of numbers that can be written as a sum and a difference of the same two quantities, $a+b$ and $a-b$, multiplies to $a^2-b^2$ instead of requiring a direct multiplication or a harder equation.

Multiply every term of the first factor by every term of the second factor.

$(a+b)(a-b)=a\cdot a - a\cdot b + b\cdot a - b\cdot b.$

The four products are $a^2$, then $-ab$, then $+ba$, then $-b^2$. Since multiplication does not care about order, $ab=ba$, so $-ab$ and $+ba$ are exact opposites and add to zero.

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

This holds for every real $a$ and $b$, since nothing in the expansion assumed any particular values.

By part A, $(a+b)(a-b)=a^2-b^2$ for every real $a$ and $b$. Ask when this could also equal $a^2+b^2$.

$a^2-b^2=a^2+b^2 \implies -b^2=b^2 \implies 2b^2=0 \implies b=0.$

So the two expressions agree only in the single case $b=0$, where both sides equal $a^2$. For every nonzero real $b$, $(a+b)(a-b)$ produces $a^2-b^2$, not $a^2+b^2$, so this particular pair of factors can never build a sum of squares.

Part A establishes a fact for every real $a,b$: $a^2-b^2$ always equals $(a+b)(a-b)$, so a difference of squares always factors this way.

Part B establishes that the SAME product can never equal $a^2+b^2$ once $b\neq0$, so this natural candidate factorization fails for a sum of squares.

$a^2+b^2 \neq (a+b)(a-b) \ \text{whenever} \ b\neq0.$

Saying a sum of squares does not factor, with no domain named, is incomplete: what part B actually rules out is a REAL linear factorization built from these two factors. The claim is really that it does not factor over the real numbers, and a later, larger number system is free to change that, without making this claim about the reals false.

Read $16x^4-81$ as a difference of two perfect squares.

$16x^4-81=(4x^2)^2-9^2=(4x^2+9)(4x^2-9).$

Inspect each new factor. The factor $4x^2-9$ is itself a difference of squares, so factor it again.

$4x^2-9=(2x)^2-3^2=(2x+3)(2x-3).$

The other factor, $4x^2+9$, is a sum of squares and does not factor over the real numbers, so the work stops there. Assembling the pieces:

$16x^4-81=(4x^2+9)(2x+3)(2x-3).$

Neither $3x^4$ nor $243$ is a perfect square on its own, so the difference-of-squares pattern does not apply yet. Pull out the common factor $3$ first.

$3x^4-243=3\left(x^4-81\right).$

What is left inside is a difference of squares.

$x^4-81=(x^2+9)(x^2-9).$

The factor $x^2-9$ is itself a difference of squares, so factor it again, while the factor $x^2+9$ is a sum of squares and does not factor over the real numbers.

$x^2-9=(x+3)(x-3).$

Keeping the $3$ out front, the complete factorization is

$3x^4-243=3(x^2+9)(x+3)(x-3).$

The habit has a fixed order: remove any common factor first, and only then look for, and re-check for, a difference of squares.

In part B, trying the pattern on the original expression before pulling out its common factor fails immediately, because neither term is a perfect square by itself: there is no square root to take and no pattern to apply yet. The common factor is not an optional cleanup step; it is what exposes the pattern in the first place, visible only once it is gone:

$3x^4-243=3\left(x^4-81\right)=3(x^2+9)(x+3)(x-3).$

Expand the proposed right side and compare it term by term with the left side.

$(8x-5)^2=(8x)^2-2(8x)(5)+5^2=64x^2-80x+25.$

This has a middle term, $-80x$, that $64x^2-25$ does not have, so the claim is false. A difference of squares needs two factors with OPPOSITE signs, not the same sign squared.

$64x^2-25=(8x)^2-5^2=(8x+5)(8x-5).$

Expand the proposed right side.

$(4x+5)(4x-5)=(4x)^2-5^2=16x^2-25.$

That is not $4x^2-25$: the coefficient came out wrong. The mistake is in the square root taken for the first term. Since $(2x)^2=4x^2$, the correct root of $4x^2$ is $2x$, not $4x$.

$4x^2-25=(2x)^2-5^2=(2x+5)(2x-5).$

The two claims fail for genuinely different reasons, and naming the reason in general terms is what makes it useful on a NEW problem, not just this one.

The first claim's error is a SIGN error: squaring two factors of the same sign, $(A-B)^2$, always leaves a middle term, $-2AB$, that a difference of squares never has. This mistake will show up on any problem where both factors are given the same sign.

The second claim's error is a ROOT error: the square root of a coefficient-bearing term was read off incorrectly. This mistake will show up on any problem where the root of a coefficient is guessed rather than computed.

Expanding catches both because it is unconditional: a genuine difference of squares always has the form $A^2-B^2$ with no middle term at all,

$(A+B)(A-B)=A^2-B^2,$

so any surviving middle term, or any constant that does not match, proves the claim wrong, regardless of which of the two mistakes produced it.