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Factoring Harder 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 direct factorization

Factor $9x^2-9x+2$ .
1. Hint 1
  The leading coefficient no longer stays out of the way: it splits across the two factors, so both factor pairs of $9$ and of $2$ are in play.
2. Hint 2
  Work out the product of the leading and constant coefficients, then look for two numbers with that product and with sum $-9$ .
3. Hint 3
  Both of those numbers are negative, since their product is positive and their sum is not. Build the candidate whose cross products are those two numbers, then expand to check.
Answer

$(3x-1)(3x-2)$ .

Full solution

The product of the leading and constant coefficients is
$9 \cdot 2 = 18,$
and the middle coefficient is $-9$ .
So the middle term splits into two numbers with product $18$ and sum $-9$ .
Both are negative, because a positive product with a negative sum forces two negative numbers.
The pair is $-3$ and $-6$ , since $-3+(-6)=-9$ and
$(-3)(-6)=18.$

Those two numbers are the cross products to aim for.
Splitting $9$ as $3 \cdot 3$ , pairing $3x$ with $-1$ gives a $-3x$ and pairing $3x$ with $-2$ gives a $-6x$ , which suggests $(3x-1)(3x-2)$ .
Expanding checks it:
$(3x-1)(3x-2)=9x^2-9x+2.$
The cross products $-3x$ and $-6x$ add to $-9x$ , so the factorization is correct.

Answer

$(3x-1)(3x-2)$ .

Key idea

The product of the leading and constant coefficients, together with the middle coefficient, tells you which two numbers the middle term splits into.
Problem 1: Mark complete
Problem 2 A split record

A middle-term split for $4x^2+bx-9$ uses the terms $12x$ and $-3x$ . Find $b$ .
1. Hint 1
  The split terms must recombine into the original middle term.
2. Hint 2
  Add their coefficients, and check that their product agrees with the leading coefficient times the constant.
Answer

$b=9$ .

Full solution

The coefficients combine as
$12+(-3)=9,$
so $b=9$ .
They also pass the product check because $12(-3)=-36$ and $4(-9)=-36$ .
Pairing $4x$ with $3$ gives the $12x$ and pairing $x$ with $-3$ gives the $-3x$ , which builds $(4x-3)(x+3)$ .
Expanding that product gives $4x^2+12x-3x-9$ , which collects to $4x^2+9x-9$ .

Answer

$b=9$ .

Key idea

A valid middle-term split has coefficients that add to the middle coefficient and multiply to the leading coefficient times the constant.
Problem 2: Mark complete
Problem 3 Leading factor choices

Find all positive integer pairs $(p,q)$ for which $(px+1)(qx+2)$ has leading coefficient $14$ and middle coefficient $11$ .
1. Hint 1
  Expand the product first and see which coefficient each of $p$ and $q$ feeds.
2. Hint 2
  Matching the two expansions gives $pq=14$ and $2p+q=11$ , so test the positive divisor choices for $p$ in turn.
Answer

$(p,q)=(2,7)$ .

Full solution

Expanding gives
$pq\,x^2+(2p+q)x+2,$
so the conditions are $pq=14$ and $2p+q=11$ .
Because $p$ divides $14$ , only four positive pairs are available.

The pairs $(1,14)$ , $(2,7)$ , $(7,2)$ , and $(14,1)$ give middle coefficients $16$ , $11$ , $16$ , and $29$ .
Only $(2,7)$ meets the requirement.
Expanding $(2x+1)(7x+2)$ confirms it, since the product is
$14x^2+11x+2.$

Answer

$(p,q)=(2,7)$ .

Key idea

Splitting the leading coefficient is constrained by the required cross-product sum.
Problem 3: Mark complete
Problem 4 A revenue target

A stall sells $24-5p$ kilograms of apples in a day when the price is $p$ dollars per kilogram, and it sets only prices with $0<p<4.8$ . The day's revenue is the price times the number of kilograms sold. Find every price at which the revenue is exactly $16$ dollars, and give the factored form you used.
1. Hint 1
  Multiplying the price by the number of kilograms turns the revenue condition into a quadratic equation, once every term sits on one side of it.
2. Hint 2
  In standard form the equation reads $5p^2-24p+16=0$ , so look for two numbers with product $80$ and sum $-24$ .
3. Hint 3
  Both of those numbers are negative. Build the candidate factors whose cross products are those numbers, then test each zero against the allowed range.
Answer

$p=\frac45$ dollars per kilogram, that is $0.80$ , and $p=4$ dollars per kilogram; from $(5p-4)(p-4)=0$ .

Full solution

The revenue is the price times the kilograms sold, so the condition is
$p(24-5p)=16.$
Expanding the left side gives
$24p-5p^2=16.$
Collecting every term on one side, so that the leading coefficient stays positive, gives
$5p^2-24p+16=0.$

The product of the leading and constant coefficients is $5 \cdot 16=80$ , and the middle coefficient is $-24$ , so the middle term splits into two numbers with product $80$ and sum $-24$ .
Both are negative, because a positive product with a negative sum forces two negative numbers, and the pair is $-20$ and $-4$ .
Pairing $5p$ with $-4$ gives a $-20p$ and pairing $p$ with $-4$ gives a $-4p$ , which suggests $(5p-4)(p-4)$ .
Expanding checks it:
$(5p-4)(p-4)=5p^2-24p+16.$

Setting each factor to zero gives $p=\frac45$ and $p=4$ , and both lie strictly between $0$ and $4.8$ , so the stall could set either price.
At $\frac45$ dollars per kilogram it sells $20$ kilograms and takes $16$ dollars, and at $4$ dollars per kilogram it sells $4$ kilograms and takes $16$ dollars.

Answer

$p=\frac45$ dollars per kilogram, that is $0.80$ , and $p=4$ dollars per kilogram; from $(5p-4)(p-4)=0$ .

Key idea

Both zeros of a factored quadratic deserve a check against the range the situation allows, and a fractional one can be perfectly admissible.
Problem 4: Mark complete
Problem 5 A non-monic equation

Find every real $t$ for which $12t^2+6t=18$ , and give the factorization you used.
1. Hint 1
  Place all terms on one side so that a zero stands alone, then look at the three coefficients together before factoring.
2. Hint 2
  All three share a numerical factor, and removing it shrinks the leading coefficient you have to break apart.
3. Hint 3
  Use the product of the simplified leading and constant coefficients to guide the two cross products.
Answer

$t=1$ or $t=-\frac32$ , also written $-1.5$ ; factorization $6(2t+3)(t-1)=0$ , or equivalently $(2t+3)(t-1)=0$ .

Full solution

The equation becomes
$12t^2+6t-18=0.$
Dividing by $6$ gives
$2t^2+t-3=0.$
The product $-6$ and sum $1$ suggest $3$ and $-2$ .
Checking binomials gives
$(2t+3)(t-1)=0.$
Restoring the common factor, the original left side factors completely as $6(2t+3)(t-1)$ , which gives the same equation.
Thus $t=-3/2$ or $t=1$ .
In the original left side, these give $27-9=18$ and $12+6=18$ .

Answer

$t=1$ or $t=-\frac32$ , also written $-1.5$ ; factorization $6(2t+3)(t-1)=0$ , or equivalently $(2t+3)(t-1)=0$ .

Key idea

Remove a common factor first to reduce the search for non-monic factors.
Problem 5: Mark complete
Problem 6 An assembly constraint

An assembly setting $v$ satisfies $8v^2-14v+3=0$ and $0<v<1$ . Find every algebraic root and the permitted setting.
1. Hint 1
  The product of the leading and constant coefficients guides a split of the middle coefficient.
2. Hint 2
  Seek two negative integers with product $24$ and sum $-14$ , then check candidate factors.
Answer

Algebraic roots: $\frac14,\frac32$ ; permitted setting: $v=\frac14$ .

Full solution

Here $ac=24$ .
The split pair $-12,-2$ has sum $-14$ .
The factors
$(4v-1)(2v-3)$
have cross terms $-12v$ and $-2v$ , and expand to the given trinomial.
Their zeros are $v=1/4$ and $v=3/2$ .
Only $1/4$ is between zero and one.
Substitution gives
$\frac12-\frac72+3=0.$

Answer

Algebraic roots: $\frac14,\frac32$ ; permitted setting: $v=\frac14$ .

Key idea

A root comes out as a fraction when a factor's leading coefficient does not divide its constant term.
Problem 6: Mark complete
Problem 7 An integer pair search

Decide whether $7x^2+5x+2$ factors over the integers, listing every integer pair you test.
1. Hint 1
  An integer factorization would make the middle term split in a particular way, so decide what the two split numbers would have to satisfy.
2. Hint 2
  Work out the product of the leading and constant coefficients, then list every integer pair with that product and compare each sum with $5$ .
3. Hint 3
  A pair of negative numbers has a positive product but a negative sum, so decide whether any negative pair could reach $5$ .
Answer

It does not factor: no integer pair has product $14$ and sum $5$ . The pairs are $1$ and $14$ (sum $15$ ), $2$ and $7$ (sum $9$ ), $-1$ and $-14$ (sum $-15$ ), and $-2$ and $-7$ (sum $-9$ ), so $7x^2+5x+2$ is irreducible over the integers.

Full solution

An integer factorization $(px+r)(qx+s)$ would make the middle coefficient the sum of two cross products whose product is the leading coefficient times the constant.
Here that product is
$7 \cdot 2 = 14.$
So the search is for two integers with product $14$ and sum $5$ .

The positive pairs with product $14$ are $1$ and $14$ , of sum $15$ , and $2$ and $7$ , of sum $9$ .
The only other integer pairs are the negatives of those two, $-1$ and $-14$ , of sum $-15$ , and $-2$ and $-7$ , of sum $-9$ .
Every pair has been tested and none has sum $5$ .

So no integer pair splits the middle term, and $7x^2+5x+2$ is irreducible over the integers.
Reporting that the search failed is the correct answer here; forcing a factorization would only produce a wrong one.

Answer

It does not factor: no integer pair has product $14$ and sum $5$ . The pairs are $1$ and $14$ (sum $15$ ), $2$ and $7$ (sum $9$ ), $-1$ and $-14$ (sum $-15$ ), and $-2$ and $-7$ (sum $-9$ ), so $7x^2+5x+2$ is irreducible over the integers.

Key idea

When no integer pair multiplies to the leading coefficient times the constant and adds to the middle coefficient, the trinomial is irreducible over the integers, and the honest report is that it does not factor.
Problem 7: Mark complete
Problem 8 A swapped pair of constants

Kai says that swapping the constants in $(2x+3)(5x+1)$ preserves the quadratic because the leading and constant terms do not change. Is the claim correct? Explain by comparing the middle terms.
1. Hint 1
  The middle term is built from the two cross products, so work those out before judging the claim.
2. Hint 2
  Compare the cross products before and after interchanging $3$ and $1$ .
Answer

No; the middle coefficients are $17$ and $11$ .

Full solution

The original cross products are $2x$ and $15x$ , giving
$2x+15x=17x.$
After the swap, $(2x+1)(5x+3)$ has cross products $6x$ and $5x$ , giving
$6x+5x=11x.$
Although both products have leading term $10x^2$ and constant $3$ , their middle terms differ.

Answer

No; the middle coefficients are $17$ and $11$ .

Key idea

Candidate factors must match the middle coefficient as well as the leading and constant terms.
Problem 8: Mark complete
Problem 9 A middle-term guide

For $15x^2-4x-4$ , a student uses $-10$ and $6$ as a middle-term guide, then proposes $(5x+2)(3x-2)$ . Determine whether both the guide and the factorization are correct, and explain.
1. Hint 1
  Test the guide against both $ac$ and $b$ .
2. Hint 2
  Expand the proposed factors to compare the cross terms with the guide.
Answer

Both are correct; $(5x+2)(3x-2)=15x^2-4x-4$ .

Full solution

The guide has product $-60$ , which is $15(-4)$ , and sum $-4$ , which is the middle coefficient.
Expanding the proposed factors gives leading term $15x^2$ , cross terms $-10x$ and $6x$ , and constant $-4$ .
Thus
$(5x+2)(3x-2)=15x^2-4x-4.$
Both checks succeed.

Answer

Both are correct; $(5x+2)(3x-2)=15x^2-4x-4$ .

Key idea

A valid middle-term guide becomes a checked factorization when its two terms appear as the cross products.
Problem 9: Mark complete
Problem 10 An integer-root prediction

Rae claims that if a quadratic has integer coefficients, every real root is an integer. Decide whether the claim is correct. If it is not, construct a quadratic with integer coefficients that factors over the integers and has a non-integer root, verify that root by substitution, and say why a factorization into monic factors could not have produced it.
1. Hint 1
  The claim covers every quadratic, so a single quadratic that fails it settles the matter; look among the non-monic quadratics this lesson factors.
2. Hint 2
  Think about what a factor has to look like for its own zero to be a fraction rather than a whole number.
3. Hint 3
  Multiply two binomials with integer entries, expand to show the coefficients are integers, then substitute the fractional zero into the expanded form.
Answer

The claim is incorrect: $3x^2+2x-1=(3x-1)(x+1)$ has the non-integer root $x=\frac13$ . Monic factors $x+r$ with integer $r$ have the integer zero $-r$ , so they cannot produce a fraction. Other counterexamples and valid reasons are accepted.

Full solution

A claim about every quadratic is refuted by one quadratic that fails it, so a single counterexample settles the question.
Build one from factors whose entries are integers, so that the expanded coefficients are integers too.

Take the product
$(3x-1)(x+1)=3x^2+2x-1,$
whose coefficients $3$ , $2$ and $-1$ are all integers.
The factor $3x-1$ is zero at $x=\frac13$ .

Substitution in the expanded form checks that root:
$3\left(\frac13\right)^2+\frac23-1=0.$
Since $\frac13$ is not an integer, Rae's claim is incorrect.
A factorization into monic factors could not have produced this root, because a monic factor $x+r$ with integer $r$ is zero at $x=-r$ , which is an integer.
The fraction needed a factor whose leading coefficient is not $1$ , here the $3$ in $3x-1$ ; equally, a product of two monic factors would have leading coefficient $1$ , and this quadratic has leading coefficient $3$ .

Answer

The claim is incorrect: $3x^2+2x-1=(3x-1)(x+1)$ has the non-integer root $x=\frac13$ . Monic factors $x+r$ with integer $r$ have the integer zero $-r$ , so they cannot produce a fraction. Other counterexamples and valid reasons are accepted.

Key idea

A claim about every quadratic is refuted by exhibiting a single quadratic that fails it.
Problem 10: Mark complete

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