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Squares of Binomials: 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 Squaring, then subtracting a square

Write $(2t+1)^2-(2t)^2$ as a polynomial with like terms combined.
1. Hint 1
  The subtracted term is the square of one of the two terms being squared.
2. Hint 2
  The two $4t^2$ terms cancel, so only the other two terms of the expansion survive.
Answer

$4t+1$ .

Full solution

The square of the sum is
$(2t+1)^2=4t^2+4t+1.$
Subtracting $(2t)^2=4t^2$ leaves $4t+1$ .
At $t=1$ , the original expression is $9-4=5$ and the polynomial is $4+1=5$ .

Answer

$4t+1$ .

Key idea

Subtracting the square of a binomial's first term from the square of the whole binomial leaves twice the product of the two terms and the square of the last term.
Problem 1: Mark complete
Problem 2 Factoring a trinomial completely

Factor $12x^2+60x+75$ completely.
1. Hint 1
  A trinomial should be in lowest terms before its outer terms are tested against the square pattern.
2. Hint 2
  Once a numerical factor is out in front, find the terms whose squares are the outer terms and compare twice their product with the middle term.
Answer

$3(2x+5)^2$ .

Full solution

Every coefficient is a multiple of $3$ , so pulling that factor out leaves
$3(4x^2+20x+25).$
Inside the parentheses the outer terms are squares, since $4x^2=(2x)^2$ and $25=5^2$ , and twice the product of $2x$ and $5$ is $20x$ , which matches the middle term.
That makes the trinomial in the parentheses $(2x+5)^2$ , so the complete factorization is
$3(2x+5)^2.$
The $3$ stays in front, because it is not the square of a whole number.
Expanding $3(2x+5)^2$ returns the original trinomial, since $3$ times $4x^2+20x+25$ rebuilds each coefficient.

Answer

$3(2x+5)^2$ .

Key idea

Removing a common numerical factor first can reveal a perfect square that the original coefficients hide.
Problem 2: Mark complete
Problem 3 An unknown coefficient

In the expansion of $(ax+4)^2$ , where $a$ is a number, the coefficient of $x$ is $48$ . Find the coefficient of $x^2$ .
1. Hint 1
  Of the three coefficients in the expansion, only one is built from both terms of the binomial.
2. Hint 2
  Expand with $a$ left in place, match the coefficient of $x$ against $48$ to find $a$ , then square the whole first term.
Answer

$36$ .

Full solution

Expanding with $a$ left in place gives
$a^2x^2+8ax+16.$
Matching the coefficient of $x$ gives $8a=48$ , so $a=6$ .
The coefficient of $x^2$ is $a^2$ , which is $36$ .
Squaring $6x+4$ gives $36x^2+48x+16$ , whose coefficient of $x$ is the given $48$ .

Answer

$36$ .

Key idea

The doubled middle term pins down the coefficient inside the binomial, and squaring that coefficient gives the leading one.
Problem 3: Mark complete
Problem 4 A placard order

Three identical square placards have a combined area of $12x^2-48x+48$ square cm, where $x>2$ . Find the side length of one placard as a linear expression in $x$ .
1. Hint 1
  Three identical squares share the total equally, and a side length belongs to one square, not to the total.
2. Hint 2
  Remove a numerical common factor and recognize the remaining trinomial as a square; the side length must be positive.
Answer

$2x-4$ cm, or equivalently $2(x-2)$ cm.

Full solution

One placard has area $4x^2-16x+16$ square cm.
Factoring gives
$4x^2-16x+16=4(x-2)^2.$
This is $(2x-4)^2$ .
Because $x>2$ , $2x-4$ is positive and is the side length.
Three copies of its square reproduce the given combined area.

Answer

$2x-4$ cm, or equivalently $2(x-2)$ cm.

Key idea

Pull out a numerical common factor before testing for a perfect square, and let the stated domain choose the positive root.
Problem 4: Mark complete
Problem 5 A student's expansion

A student writes $(3y-7)^2=3y^2-42y-49$ . Identify both errors in that expansion, write the correct expansion, and use one value of $y$ to show that the student's version is wrong.
1. Hint 1
  A coefficient belongs to the term being squared, and the square of a negative number is positive.
2. Hint 2
  Expand with the difference pattern, then compare the result term by term with what the student wrote.
Answer

The coefficient $3$ was not squared, so the leading term is $9y^2$ , and the last term is $+49$ , not $-49$ ; the correct expansion is $9y^2-42y+49$ . At $y=1$ it gives $16$ , while the student's version gives $-88$ .

Full solution

Squaring $3y-7$ squares the whole first term, so the leading term is $(3y)^2=9y^2$ and not $3y^2$ : the student squared the variable but left the coefficient behind.
The last term is the square of $-7$ , which is $+49$ , so the sign of the student's $-49$ is the second error.
The middle term is correct, since
$-2(3y)(7)=-42y.$

The correct expansion is
$9y^2-42y+49.$
At $y=1$ the original expression is $(3-7)^2=16$ , and the correct expansion gives
$9-42+49=16.$
The student's version gives $3-42-49=-88$ , so it is not equal to $(3y-7)^2$ .

Answer

The coefficient $3$ was not squared, so the leading term is $9y^2$ , and the last term is $+49$ , not $-49$ ; the correct expansion is $9y^2-42y+49$ . At $y=1$ it gives $16$ , while the student's version gives $-88$ .

Key idea

Squaring a term squares its coefficient too, and squaring a negative last term leaves it positive.
Problem 5: Mark complete
Problem 6 When two expressions agree

Solve $(2x-3)^2=4x^2-27$ for $x$ .
1. Hint 1
  A squared binomial carries a linear term that the other expression does not have.
2. Hint 2
  The squared terms cancel from the equation, leaving a linear condition.
Answer

$x=3$ .

Full solution

Expanding the left side gives
$4x^2-12x+9=4x^2-27.$
Subtracting $4x^2$ and then $9$ gives
$-12x=-36.$
Thus $x=3$ .
At $x=3$ the two sides are $3^2=9$ and $36-27=9$ , confirming equality.

Answer

$x=3$ .

Key idea

An equation containing two quadratic expressions may become linear after their squared terms cancel.
Problem 6: Mark complete
Problem 7 A square's measurements

A square has area $64t^2+80t+25$ square cm and perimeter $60$ cm, where $t>0$ . Find $t$ .
1. Hint 1
  The perimeter of a square is four times its side length.
2. Hint 2
  The area is a squared linear expression, which gives the positive side length.
Answer

$t=\tfrac54$ , or $1.25$ .

Full solution

The area factors as
$(8t+5)^2.$
Since $t>0$ , the side is $8t+5$ cm.
A perimeter of $60$ cm means each side is $15$ cm, so
$8t+5=15.$
This gives
$t=\tfrac54.$
Substitution gives side $15$ cm, area $225$ square cm, and the required perimeter.

Answer

$t=\tfrac54$ , or $1.25$ .

Key idea

Recognizing an area as a square can turn a perimeter condition into a linear equation.
Problem 7: Mark complete
Problem 8 Lin's average

For real numbers $u,v$ , Lin claims that the average of $(u+v)^2$ and $(u-v)^2$ is $u^2+v^2$ . Decide whether the claim is correct and justify your answer.
1. Hint 1
  Squaring a sum and squaring a difference produce the same two outer squares.
2. Hint 2
  Their middle terms have opposite signs; track them when the two expansions are added.
Answer

The claim is correct for all real $u,v$ .

Full solution

The expansions are $u^2+2uv+v^2$ and $u^2-2uv+v^2$ .
Their sum is
$2u^2+2v^2.$
Dividing by $2$ gives $u^2+v^2$ .
No sign restriction was used, so the claim holds for every pair of real numbers.

Answer

The claim is correct for all real $u,v$ .

Key idea

Averaging squares with opposite cross terms leaves the sum of the two individual squares.
Problem 8: Mark complete
Problem 9 A size prediction

Bo claims that $(a+b)^2$ is greater than $a^2+b^2$ whenever $a$ and $b$ are nonzero real numbers. Decide whether the claim is correct, and state exactly when the first expression is greater.
1. Hint 1
  The two expressions differ only by the cross term that the expansion produces.
2. Hint 2
  The sign of the remaining product depends on the signs of $a$ and $b$ .
Answer

False; it is greater exactly when $ab>0$ , meaning $a,b$ have the same sign.

Full solution

Expanding and subtracting leaves
$D=2ab,$
where $D$ is the first expression minus the second.
Thus $D>0$ exactly when $ab>0$ .
Nonzero numbers of opposite signs give $D<0$ : for $a=1,b=-1$ , the first expression is $0$ and the second is $2$ .

Answer

False; it is greater exactly when $ab>0$ , meaning $a,b$ have the same sign.

Key idea

The cross term determines whether a squared sum is greater or smaller than the sum of the separate squares.
Problem 9: Mark complete
Problem 10 Two formula labels

Find every real $k$ for which $(kx+4)^2$ and $(kx-4)^2$ are equal for every real $x$ , and explain why no other value works.
1. Hint 1
  Compare the two expansions as polynomials in $x$ .
2. Hint 2
  The squared and constant terms agree already; equality for every input also requires the linear terms to agree.
Answer

$k=0$ .

Full solution

The expansions are $k^2x^2+8kx+16$ and $k^2x^2-8kx+16$ .
Their difference is
$16kx.$
For this to vanish at every real $x$ , it must vanish at $x=1$ , forcing $k=0$ .
With $k=0$ , both expressions are $16$ for every input, so the condition is sufficient as well.

Answer

$k=0$ .

Key idea

An identity requires matching cross terms at every input, not agreement at a single special input.
Problem 10: Mark complete

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