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The Binomial Theorem: 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: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra II. You can skip it.

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Problem 1 of 10

Problem 1 A cubic expression

Expand $(x^2+2y)^3$ fully.
1. Hint 1
  Treat $x^2$ and $2y$ as the two complete terms of the binomial.
2. Hint 2
  The cubic coefficients are $1,3,3,1$ ; each power applies to the coefficient as well as the variable.
Answer

$x^6+6x^4y+12x^2y^2+8y^3$ .

Full solution

Using $A=x^2$ and $B=2y$ , the four contributions are $A^3$ , $3A^2B$ , $3AB^2$ , and $B^3$ .
They simplify to $x^6$ , $6x^4y$ , $12x^2y^2$ , and $8y^3$ .

Thus, writing $F=(x^2+2y)^3$ ,
$F=x^6+6x^4y+12x^2y^2+8y^3$
At $x=y=1$ , both the original and expanded expressions are $27$ , providing a check.

Answer

$x^6+6x^4y+12x^2y^2+8y^3$ .

Key idea

A binomial expansion raises each complete term, including its coefficient and internal exponent.
Problem 1: Mark complete
Problem 2 A difference of powers

Expand and simplify $(x+1)^4-(x-1)^4$ .
1. Hint 1
  The two expansions share their even-power terms.
2. Hint 2
  Expand each fourth power, then subtract the entire second expansion.
Answer

$8x^3+8x$ .

Full solution

The first expansion is $x^4+4x^3+6x^2+4x+1$ .
The second is $x^4-4x^3+6x^2-4x+1$ .

Subtracting cancels the terms in $x^4$ , $x^2$ , and the constants; the others double.
$(x+1)^4-(x-1)^4=8x^3+8x$
At $x=1$ , both sides equal $16$ .

Answer

$8x^3+8x$ .

Key idea

Subtracting opposite-sign binomial expansions cancels the terms whose signs agree.
Problem 2: Mark complete
Problem 3 A squared variable

Find the coefficient of $t^7$ in $(t+t^2)^5$ .
1. Hint 1
  Each choice of $t^2$ contributes one extra power of $t$ beyond the five already present.
2. Hint 2
  The general exponent is $5+k$ when k factors contribute $t^2$ .
Answer

$10$ .

Full solution

A term taking $t^2$ from $k$ factors and $t$ from the other $5-k$ factors has exponent $5+k$ .
The target $7$ requires $k=2$ .
$\binom52=10$
Thus the term is $10t^7$ .
Factoring out $t^5$ from $t^5(1+t)^5$ gives the same coefficient.

Answer

$10$ .

Key idea

When both binomial terms contain the variable, combine their exponents before choosing the term index.
Problem 3: Mark complete
Problem 4 A paired product

Expand $(x+1)^3(x-1)^3$ as a polynomial in descending powers of $x$ .
1. Hint 1
  Equal powers can be combined into one power of the product.
2. Hint 2
  The product of $x+1$ and $x-1$ is a difference of squares.
Answer

$x^6-3x^4+3x^2-1$ .

Full solution

Combine the two cubic powers.
$(x+1)^3(x-1)^3=(x^2-1)^3$
Now expand the cube using $A=x^2$ and $B=-1$ .

The result is $x^6-3x^4+3x^2-1$ .
Substituting $x=1$ gives zero on both sides, and $x=0$ gives $-1$ on both sides.

Answer

$x^6-3x^4+3x^2-1$ .

Key idea

Combining a conjugate pair before expanding can reduce the number of binomial terms to handle.
Problem 4: Mark complete
Problem 5 A coefficient balance

For a positive integer $n\ge2$ , the sum of the coefficients of $x$ and $x^2$ in $(1+x)^n$ is $66$ . Find $n$ and verify both coefficients.
1. Hint 1
  The coefficients are the counts of choosing one or two x terms from the factors.
2. Hint 2
  Write them as $n$ and $n(n-1)/2$ , then solve their sum equation.
Answer

$n=11$ ; the coefficients are $11$ and $55$ .

Full solution

The coefficient condition is
$n+n(n-1)/2=66.$
Multiplying by $2$ gives
$n^2+n-132=0$
so
$(n-11)(n+12)=0.$

The allowed positive integer is $n=11$ .
The coefficients are $\binom{11}1=11$ and $\binom{11}2=55$ , and their sum is $66$ .
The negative root is not an allowable row number.

Answer

$n=11$ ; the coefficients are $11$ and $55$ .

Key idea

Factorial coefficient formulas can turn information about an expansion into an equation for its exponent.
Problem 5: Mark complete
Problem 6 Pascal's rule in reverse

Three consecutive entries in one row of Pascal's triangle are $16$ , $B$ and $560$ . The two entries directly below them, between each neighboring pair, are $136$ and $680$ . Find $B$ in two ways, then find the entry directly between and below $136$ and $680$ in the following row.
1. Hint 1
  Each interior entry is the sum of the two entries directly above it.
2. Hint 2
  Undo that addition twice to find B, then apply it forward once.
Answer

$B=120$ ; the next entry is $816$ .

Full solution

Since $136=16+B$ , $B=120$ .
Independently, $680=B+560$ also gives $B=120$ , so the two readings agree.

The entry below $136$ and $680$ is their sum, $816$ .
These are $\binom{16}1$ , $\binom{16}2$ and $\binom{16}3$ , then $\binom{17}2$ , $\binom{17}3$ and $\binom{18}3$ .

Answer

$B=120$ ; the next entry is $816$ .

Key idea

Pascal's addition rule can be run backward to recover a missing entry and forward to build the next row.
Problem 6: Mark complete
Problem 7 An unknown constant

Let $c$ be a nonzero real number. In $(x+c)^5$ , the coefficient of $x^3$ is the negative of the coefficient of $x^2$ . Find $c$ and the two coefficients.
1. Hint 1
  Write the two coefficients using the number of constant terms chosen.
2. Hint 2
  The assumption $c\ne0$ permits canceling a power of c in the coefficient equation.
Answer

$c=-1$ ; coefficient of $x^3$ is $10$ , and coefficient of $x^2$ is $-10$ .

Full solution

The coefficients are $\binom52c^2=10c^2$ and
$\binom53c^3=10c^3.$
The condition gives
$10c^2=-10c^3$

Since $10c^2\ne0$ , division gives $1=-c$ , so $c=-1$ .
Substitution gives $10$ and $-10$ , which have the required relationship.

Answer

$c=-1$ ; coefficient of $x^3$ is $10$ , and coefficient of $x^2$ is $-10$ .

Key idea

Comparing binomial coefficients can determine a parameter when any canceled factors are known to be nonzero.
Problem 7: Mark complete
Problem 8 A counting statement

A student says the coefficient of $x^{n-k}y^k$ in $(x+y)^n$ counts the ways to choose which $k$ of the $n$ factors supply $y$ , for whole numbers $0\le k\le n$ . Is this correct? State the Binomial Theorem and explain the count.
1. Hint 1
  Each product formed by distribution selects one term from each factor.
2. Hint 2
  Once the factors supplying y are chosen, all the other factors supply x.
Answer

Yes; $(x+y)^n=\sum_{k=0}^n\binom nk x^{n-k}y^k$ .

Full solution

Choosing $y$ from $k$ factors leaves $n-k$ factors supplying $x$ , so each such selection produces $x^{n-k}y^k$ .
The number of selections is $\binom nk$ .

Adding the contributions for all possible $k$ gives
$(x+y)^n=\sum_{k=0}^n\binom nk x^{n-k}y^k$
The allowed indices run from zero to $n$ , so all possible selections are counted.

Answer

Yes; $(x+y)^n=\sum_{k=0}^n\binom nk x^{n-k}y^k$ .

Key idea

A binomial coefficient counts the factor selections that contribute to one collected term.
Problem 8: Mark complete
Problem 9 Two labelings

Two students find the coefficient of $x^{11}$ in $(x^2+x^5)^4$ . One uses $k=1$ and the other uses $k=3$ in $\binom4k A^{4-k}B^k$ . Can both be correct? State each student's choice of $A$ and $B$ , and find the coefficient.
1. Hint 1
  The index k counts how many factors supply the second term, B.
2. Hint 2
  Swapping which term is called A and which is called B changes the k that gives the same power.
Answer

Yes: $A=x^2$ , $B=x^5$ with $k=1$ , or $A=x^5$ , $B=x^2$ with $k=3$ ; the coefficient is $4$ .

Full solution

With $A=x^2$ and $B=x^5$ , a term has exponent $2(4-k)+5k=8+3k$ , which is $11$ at $k=1$ , with coefficient
$\binom41=4.$

With $A=x^5$ and $B=x^2$ , the exponent is $5(4-k)+2k=20-3k$ , which is $11$ at $k=3$ , with coefficient
$\binom43=4.$
Both describe the same term, choosing $x^5$ from one factor and $x^2$ from three.

Answer

Yes: $A=x^2$ , $B=x^5$ with $k=1$ , or $A=x^5$ , $B=x^2$ with $k=3$ ; the coefficient is $4$ .

Key idea

Swapping which term is labeled A and which B replaces the index k by n minus k and leaves the term itself unchanged.
Problem 9: Mark complete
Problem 10 Signs of whole terms

Let $u<0$ and $v>0$ . Without expanding fully, determine the sign of every term in $(2u-v)^5$ , and explain how this fits with its alternating coefficients.
1. Hint 1
  The sign of a coefficient and the sign of a whole term are different questions.
2. Hint 2
  Count the negative factors in each term, from both powered factors.
Answer

Every one of the six terms is negative; the coefficients alternate in sign, and so do the powers of $u$ .

Full solution

Each term is $\binom5k(2u)^{5-k}(-v)^k$ for
$k=0,1,\ldots,5.$
Since $2u<0$ and $-v<0$ , the term is a positive count times a product of $5-k$ plus $k$ , that is five, negative factors, so every term is negative.

The coefficients $\binom5k2^{5-k}(-1)^k$ alternate in sign with $k$ , and the monomials $u^{5-k}v^k$ alternate in sign too, because $u<0$ .
The two alternations cancel, leaving every term negative.

Answer

Every one of the six terms is negative; the coefficients alternate in sign, and so do the powers of $u$ .

Key idea

A term's sign depends on both its coefficient and the signs of the variables, so alternating coefficients need not give alternating terms.
Problem 10: Mark complete

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