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Factoring Quadratics: 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 entry

Complete the identity $(x+\square)(x+6)=x^2+10x+24$ with an integer.
1. Hint 1
  The constant term comes from multiplying the two constant entries.
2. Hint 2
  The missing entry must also make the two constants add to the coefficient of $x$ .
Answer

$4$ .

Full solution

Let the missing entry be $p$ .
The constant terms require
$6p=24,$
so $p=4$ .
The middle coefficient checks because $4+6=10$ .
Thus $(x+4)(x+6)$ expands to the required trinomial.

Answer

$4$ .

Key idea

Both the constant term and the middle coefficient constrain a missing factor entry.
Problem 1: Mark complete
Problem 2 Factor and check

Factor $x^2-4x-32$ , then check your factors by expanding them.
1. Hint 1
  The sign of the constant term tells you whether the two entries share a sign.
2. Hint 2
  List the pairs of whole numbers whose product is $32$ , then attach signs so that the sum comes to $-4$ .
Answer

$(x-8)(x+4)$ .

Full solution

The constant $-32$ is negative, so the two entries have opposite signs.
Their sum is $-4$ , which is negative, so the entry of greater size is the negative one.
The size pairs with product $32$ are $1$ and $32$ , then $2$ and $16$ , then $4$ and $8$ , and placing the minus sign on the larger member gives the sums $-31$ , $-14$ and $-4$ .
Only the last pair fits, so the entries are $-8$ and $4$ .
$x^2-4x-32=(x-8)(x+4)$
Expanding rebuilds the original, since
$x^2+4x-8x-32=x^2-4x-32.$

Answer

$(x-8)(x+4)$ .

Key idea

Read the sign of the constant term first, then let the middle coefficient decide where the minus sign goes.
Problem 2: Mark complete
Problem 3 Solve and check

Solve $x^2+9x+14=0$ by factoring, then check one of your roots by substitution.
1. Hint 1
  Both the constant term and the middle coefficient are positive here, which fixes the sign of each entry.
2. Hint 2
  Once the left side is written as a product, use the fact that a product is zero only when one of its factors is zero.
Answer

$x=-2$ or $x=-7$ .

Full solution

The constant $14$ is positive, so the two entries share a sign, and the sum $9$ is positive, so both are positive.
The positive pairs with product $14$ are $1$ and $14$ , with sum $15$ , and $2$ and $7$ , with sum $9$ .
The second pair fits.
$x^2+9x+14=(x+2)(x+7)$
A product is zero only when one factor is zero, so $x+2=0$ or $x+7=0$ .
The roots are $x=-2$ and $x=-7$ , each the opposite sign of the number inside its factor.
Substituting $x=-2$ confirms the first root, since $4-18+14=0$ .

Answer

$x=-2$ or $x=-7$ .

Key idea

Once one side is a product and the other is zero, each factor set to zero contributes one root.
Problem 3: Mark complete
Problem 4 Two ages from a product

Mara is $3$ years older than Kip, and the product of their ages in years is $88$ . Find both ages, and say why the other root of your equation is not used.
1. Hint 1
  Name Kip's age with a letter, then write Mara's age using that same letter.
2. Hint 2
  Multiply the two ages, move every term to one side, and factor the trinomial that results.
Answer

Kip is $8$ years old and Mara is $11$ years old; the negative root is rejected: an age is positive.

Full solution

Let Kip's age be $k$ years, so Mara's age is $k+3$ years.
Their product is $88$ .
$k(k+3)=88$
Expanding and moving $88$ across gives a trinomial equal to zero.
$k^2+3k-88=0$
Two entries with product $-88$ and sum $3$ are $11$ and $-8$ .
$(k+11)(k-8)=0$
The roots are $k=-11$ and $k=8$ .
An age is positive, so the root $-11$ is not used, which leaves Kip at $8$ and Mara at $11$ .
Their product is $8 \cdot 11$ , which is $88$ as required.

Answer

Kip is $8$ years old and Mara is $11$ years old; the negative root is rejected: an age is positive.

Key idea

The context can rule out one algebraic root, so test every root against the conditions of the problem.
Problem 4: Mark complete
Problem 5 A printing total

A printer makes $n$ identical sheets, where $n$ is a whole number. The total charge is $3n^2+12n$ dollars. The total is $180$ dollars. Find the number of sheets.
1. Hint 1
  Put the charge equation in a form with zero on one side.
2. Hint 2
  Remove the common numerical factor before looking for two integer factor entries.
Answer

$6$ sheets.

Full solution

The charge equation gives
$3n^2+12n-180=0.$
Dividing by $3$ yields
$n^2+4n-60=0.$
The entries $10$ and $-6$ give
$(n+10)(n-6)=0.$
Thus $n=-10$ or $n=6$ .
Only $n=6$ satisfies $n \ge 0$ , and
$3(6)^2+12(6)=180.$

Answer

$6$ sheets.

Key idea

Removing a common numerical factor makes a quadratic easier to factor without changing its roots.
Problem 5: Mark complete
Problem 6 Two payment rules

For a whole number of units $u$ , one plan charges $u^2+2u$ dollars and another charges $9u$ dollars. Find every value of $u$ for which the charges agree.
1. Hint 1
  Two charges agree exactly at the values of $u$ that satisfy one equation, so work with that equation rather than testing values one at a time.
2. Hint 2
  Move every term to one side, then factor the left side completely rather than dividing through by anything.
Answer

$u=0$ or $u=7$ .

Full solution

Equality of the charges gives
$u^2+2u=9u.$
Moving terms and factoring produces
$u^2-7u=0,$ $u(u-7)=0.$
The values are $u=0$ and $u=7$ .
At zero both charges are zero; at seven both are $63$ dollars.

Answer

$u=0$ or $u=7$ .

Key idea

Factoring out a shared variable preserves the zero root that division by the variable would lose.
Problem 6: Mark complete
Problem 7 A one-entry repair

A worksheet shows $x^2-5x-50=(x-10)(x+2)$ . Change exactly one constant entry in the two factors so that the identity becomes correct, and verify your repair by expansion.
1. Hint 1
  The two entries must satisfy the required product and the required sum at the same time.
2. Hint 2
  Keep one entry, work out what the other would have to be, then test that pair against both conditions.
Answer

Replace $2$ with $5$ : $(x-10)(x+5)$ .

Full solution

The two entries need product $-50$ and sum $-5$ , and the printed factorization fails both, since $(x-10)(x+2)$ expands to $x^2-8x-20$ .
Keeping $-10$ forces the other entry to be $5$ , since $-10$ times $5$ is $-50$ , and that pair also has the sum $-5$ .
Expansion checks the repair.
$(x-10)(x+5)=x^2-5x-50$
Keeping $2$ instead forces the other entry to be $-25$ , whose sum with $2$ is $-23$ rather than $-5$ , so changing that entry alone cannot work.

Answer

Replace $2$ with $5$ : $(x-10)(x+5)$ .

Key idea

A repair to a factorization must restore both the product and the sum of the two entries.
Problem 7: Mark complete
Problem 8 An irreducibility claim

Pat claims that for every even integer $b$ , the trinomial $x^2+bx+33$ is irreducible over the integers. Decide whether the claim is correct and justify your answer.
1. Hint 1
  A positive constant term means the two integer entries share a sign.
2. Hint 2
  List the integer pairs whose product is $33$ , find each pair's sum, then test those sums against the claim about every even $b$ .
Answer

No; for example, $x^2+14x+33=(x+3)(x+11)$ .

Full solution

The positive integer pairs with product $33$ are $1$ and $33$ , with sum $34$ , and $3$ and $11$ , with sum $14$ .
Their negative versions have sums $-34$ and $-14$ .
Each of those four sums is a value of $b$ for which the trinomial does factor.
Taking the pair $3$ and $11$ gives a factorization.
$x^2+14x+33=(x+3)(x+11)$
Since $14$ is even, the claim fails at $b=14$ , and one counterexample is enough to refute a claim made about every even $b$ .

Answer

No; for example, $x^2+14x+33=(x+3)(x+11)$ .

Key idea

A claim about a whole family of coefficients is refuted by a single integer pair that fits.
Problem 8: Mark complete
Problem 9 A coefficient family

An integer $b$ makes $x^2+bx+17$ factorable as $(x+p)(x+q)$ with integers $p,q$ . Find every possible value of $b$ and explain why no other value works.
1. Hint 1
  The constant term limits the possible integer factor entries.
2. Hint 2
  List both the positive and negative pairs with product $17$ , then find their sums.
Answer

The two values are $b=18$ and $b=-18$ ; $(1,17)$ and $(-1,-17)$ are the only integer pairs with product $17$ .

Full solution

The integer pairs with product $17$ are $(1,17)$ and $(-1,-17)$ , apart from reversing their order.
Their sums are $18$ and $-18$ .
Thus the possibilities are
$x^2+18x+17=(x+1)(x+17)$
and
$x^2-18x+17=(x-1)(x-17).$
Every other integer $b$ gives an irreducible trinomial over the integers because the complete pair list has no matching sum.

Answer

The two values are $b=18$ and $b=-18$ ; $(1,17)$ and $(-1,-17)$ are the only integer pairs with product $17$ .

Key idea

Listing every integer pair for the constant term determines exactly which middle coefficients permit integer factors.
Problem 9: Mark complete
Problem 10 A student's report

A student factors $x^2-11x+30$ as $(x-6)(x-5)$ , then reports that the roots of $x^2-11x+30=0$ are $-6$ and $-5$ . Decide whether the reported roots are correct, give the correct roots, and check one by substitution.
1. Hint 1
  A root is a value that makes the trinomial equal to zero, so any reported root can be tested by substitution.
2. Hint 2
  Set each factor equal to zero and solve it, rather than copying the numbers printed inside the factors.
Answer

The reported roots are not correct. The roots are $x=6$ and $x=5$ .

Full solution

Expanding the proposed factors rebuilds the trinomial, so the factoring itself is correct.
$(x-6)(x-5)=x^2-11x+30$
A root is a value that makes one factor zero, so solve $x-6=0$ and $x-5=0$ .
That gives $x=6$ and $x=5$ , each the opposite sign of the number printed inside its factor.
Substituting $x=6$ gives $36-66+30=0$ , and substituting $x=5$ gives $25-55+30=0$ , so both are roots.
The reported value $-6$ fails the same test, since $36+66+30=132$ , so the student copied the entries instead of solving each factor.

Answer

The reported roots are not correct. The roots are $x=6$ and $x=5$ .

Key idea

The factor $x-r$ is zero at $x=r$ , so a root has the opposite sign of the entry written in its factor.
Problem 10: Mark complete

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