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Difference of Squares: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 A missing factor

Find the linear expression that replaces the blank in $(5u+2v)\,\square=25u^2-4v^2$ for all real $u,v$ .
1. Hint 1
  Each term on the right is the square of a complete variable term.
2. Hint 2
  The given factor adds those terms; the other factor must make the cross terms cancel.
Answer

$5u-2v$ .

Full solution

The squared terms are $(5u)^2$ and $(2v)^2$ .
The required factor is $5u-2v$ , since
$(5u+2v)(5u-2v)=25u^2-4v^2$
The cross products $-10uv$ and $10uv$ cancel.

Answer

$5u-2v$ .

Key idea

A missing opposite-sign factor is recovered from the square roots of both complete terms.
Problem 1: Mark complete
Problem 2 Small factor entries

In $64x^2-c^2$ , the positive integer $c$ is odd and less than $6$ . List every resulting pair of linear factors.
1. Hint 1
  First list the positive odd possibilities for $c$ .
2. Hint 2
  The square root of the variable term is $8x$ , so pair it with each possible constant using opposite signs.
Answer

$(8x+1)(8x-1)$ , $(8x+3)(8x-3)$ , and $(8x+5)(8x-5)$ .

Full solution

The allowed values of $c$ are $1,3,5$ .
For each value,
$64x^2-c^2=(8x+c)(8x-c).$
Substituting those three constants gives the three listed products.
Their constant terms are respectively $-1,-9,-25$ , matching the subtracted squares.

Answer

$(8x+1)(8x-1)$ , $(8x+3)(8x-3)$ , and $(8x+5)(8x-5)$ .

Key idea

Restrictions on a square root of the constant term translate directly into restrictions on its opposite-sign factors.
Problem 2: Mark complete
Problem 3 A comparison offset

Subtract $(3x+2)(3x-2)$ from $9x^2+1$ and simplify.
1. Hint 1
  The two factors have matching terms and opposite signs.
2. Hint 2
  After their cross terms cancel, subtract the entire resulting expression.
Answer

$5$ .

Full solution

The product is
$(3x+2)(3x-2)=9x^2-4.$
Subtracting it from $9x^2+1$ gives $1-(-4)=5$ .
Thus the difference is independent of $x$ .

Answer

$5$ .

Key idea

Recognizing canceling cross terms can reveal that the difference between two expressions is constant.
Problem 3: Mark complete
Problem 4 A sum and a square difference

Two real numbers $a$ and $b$ , in that order, have sum $14$ , and the square of the first minus the square of the second is $42$ . Find the ordered pair, showing a product expression for the difference of their squares.
1. Hint 1
  The difference of the squares is the product of the sum and the difference.
2. Hint 2
  The known sum turns that product equation into a linear equation for the difference.
Answer

$(a,b)=(\frac{17}{2},\frac{11}{2})$ , or $(8.5,5.5)$ ; product expression $(a+b)(a-b)$ .

Full solution

Their square difference is
$(a+b)(a-b)=42.$
Using $a+b=14$ gives $14(a-b)=42$ , so $a-b=3$ .
Combining sum $14$ and difference $3$ gives $a=17/2$ and $b=11/2$ .
Their sum is $14$ and their square difference is
$(289-121)/4=42.$

Answer

$(a,b)=(\frac{17}{2},\frac{11}{2})$ , or $(8.5,5.5)$ ; product expression $(a+b)(a-b)$ .

Key idea

A known sum converts a difference of squares into a condition on the difference of the two numbers.
Problem 4: Mark complete
Problem 5 A quotient to simplify

For $x\ne-13$ , let $Q=\frac{(x^2+169)(x^2-169)}{x+13}$ . Write $Q$ as a product of integer-coefficient factors that cannot be factored further over the real numbers, and find every permitted real $x$ for which $Q=0$ .
1. Hint 1
  Inspect the quadratic factors before canceling anything.
2. Hint 2
  Factor the difference of squares, cancel the nonzero denominator factor, and consider which remaining factors can be zero.
Answer

$Q=(x^2+169)(x-13)$ ; zero at $x=13$ ; $x=-13$ remains excluded.

Full solution

The difference factors as
$x^2-169=(x+13)(x-13).$
Since $x+13\ne0$ , canceling that factor gives
$Q=(x^2+169)(x-13).$
The sum $x^2+169$ is positive for every real $x$ and does not factor further over the real numbers.
Thus $Q=0$ exactly when $x=13$ , which is permitted.
The original denominator at $13$ is $26$ , which is not zero, so that value is permitted.

Answer

$Q=(x^2+169)(x-13)$ ; zero at $x=13$ ; $x=-13$ remains excluded.

Key idea

Factor before canceling, and retain every exclusion imposed by the original denominator.
Problem 5: Mark complete
Problem 6 A sum beside a difference

Consider $P=3x^2+48y^2$ and $Q=3x^4-48y^4$ . Write each as a completely factored product with integer coefficients, and justify in each case that no further factoring is possible over the real numbers.
1. Hint 1
  Each expression has the same numerical factor in both of its terms; look at what is left once that factor is gone.
2. Hint 2
  Inside each pair of parentheses, decide whether the two perfect squares are added or subtracted, taking each coefficient with its variable.
3. Hint 3
  A fourth power is the square of a square, so a first pass on $Q$ can leave a factor that is itself a difference of squares.
Answer

$P=3(x^2+16y^2)$ and $Q=3(x^2+4y^2)(x+2y)(x-2y)$ .

Full solution

Both expressions share the numerical common factor $3$ , which gives
$P=3(x^2+16y^2)$
and
$Q=3(x^4-16y^4).$
In $P$ the two perfect squares inside are $(x)^2$ and $(4y)^2$ , since $16y^2=(4y)^2$ , and they are added.
A sum of squares has no factorization over the real numbers, so $P$ stops there.

Inside $Q$ the two perfect squares are $(x^2)^2$ and $(4y^2)^2$ , since $16y^4=(4y^2)^2$ , and they are subtracted.
One pass with roots $x^2$ and $4y^2$ gives
$Q=3(x^2+4y^2)(x^2-4y^2).$
That is not yet complete, because $x^2-4y^2$ is again a difference of squares, with roots $x$ and $2y$ :
$x^2-4y^2=(x+2y)(x-2y).$
Assembling the pieces gives
$Q=3(x^2+4y^2)(x+2y)(x-2y)$
as the complete factorization.
The factor $x^2+4y^2$ is a sum of squares and the last two factors are linear, so the work stops.
Expanding the final pair returns $x^2-4y^2$ , because the cross terms $-2xy$ and $2xy$ cancel.

Answer

$P=3(x^2+16y^2)$ and $Q=3(x^2+4y^2)(x+2y)(x-2y)$ .

Key idea

Pull out the common factor first, then keep checking each new factor: a sum of squares stops, while a difference of squares comes apart again.
Problem 6: Mark complete
Problem 7 A balanced pair

Find every real $x$ for which $(4x+1)^2-(x+3)^2=0$ . Give a product form for the left side.
1. Hint 1
  Treat each full binomial as one squared quantity.
2. Hint 2
  Add the two quantities for one factor and subtract them for the other, then simplify each factor.
Answer

Product: $(5x+4)(3x-2)$ ; solutions $-\frac45$ and $\frac23$ .

Full solution

The sum of the squared quantities is $5x+4$ , and their difference is $3x-2$ .
Hence the equation is
$(5x+4)(3x-2)=0.$
The zeros are $x=-4/5$ and $x=2/3$ .
At $x=2/3$ both original quantities are $11/3$ ; at $x=-4/5$ they are $-11/5$ and $11/5$ , so their squares agree in both cases.

Answer

Product: $(5x+4)(3x-2)$ ; solutions $-\frac45$ and $\frac23$ .

Key idea

A difference of squares can be factored even when each squared quantity is itself a binomial.
Problem 7: Mark complete
Problem 8 Reversing both factors

Dev says that $(7-x)(7+x)$ equals $(x-7)(x+7)$ for every real $x$ , since the terms were only reordered. Decide whether this is correct and find every input where the two products do agree.
1. Hint 1
  Reversing a subtraction changes its sign.
2. Hint 2
  Write each product as a difference of squares, then solve the equality.
Answer

False; they agree exactly at $x=-7$ and $x=7$ .

Full solution

The first product is $49-x^2$ , and the second is $x^2-49$ .
Equality requires
$49-x^2=x^2-49,$
which gives $x^2=49$ .
Thus the agreeing inputs are $-7$ and $7$ , where both products are zero.
For example, at $x=0$ the products are $49$ and $-49$ , so they are not identical.

Answer

False; they agree exactly at $x=-7$ and $x=7$ .

Key idea

Changing the order of terms in a difference negates it, unlike changing the order of factors in a product.
Problem 8: Mark complete
Problem 9 Signs of two factors

For real numbers $p,q$ , suppose $p^2-q^2>0$ . Mei claims that $p+q$ and $p-q$ must have the same sign. Is the claim correct? Justify your answer.
1. Hint 1
  Express the given positive quantity as a product.
2. Hint 2
  Ask what the signs of the two factors can be if their product is positive, and whether either factor may be zero.
Answer

The claim is correct.

Full solution

The given condition can be rewritten as
$(p+q)(p-q)>0.$
Neither factor can be zero, since that would make the product zero.
Opposite signs would make the product negative, so both factors are positive or both are negative.
This proves Mei is correct.

Answer

The claim is correct.

Key idea

A positive difference of squares imposes a same-sign condition on its two linear factors.
Problem 9: Mark complete
Problem 10 Two related products

Let $a,b$ be positive real constants. Find every real $x$ for which $(ax+b)(ax-b)=(ax-b)^2$ , and explain why canceling $ax-b$ at the start would be misleading.
1. Hint 1
  Bring the products together while retaining the factor they share.
2. Hint 2
  The difference of the other factors is a nonzero constant because $b>0$ .
Answer

$x=\frac ba$ ; cancellation would remove this solution.

Full solution

Both sides share the factor $ax-b$ , so subtract the right side and take that factor out.
The second factor is then $(ax+b)-(ax-b)$ , which simplifies to $2b$ , giving
$2b(ax-b)=0.$
Since $b>0$ , the factor $2b$ is nonzero, so $ax-b=0$ .
Since $a>0$ , division by $a$ is valid and gives $x=b/a$ .
Both original products are then zero.
Canceling $ax-b$ would exclude exactly this solution and leave the impossible statement $2b=0$ .

Answer

$x=\frac ba$ ; cancellation would remove this solution.

Key idea

Retaining a shared factor protects a solution at which both products vanish.
Problem 10: Mark complete

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