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Solving Quadratics by Factoring: 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 Two shifted expressions

Solve $(x+2)(x-3)=2(x+2)$ over the real numbers.
1. Hint 1
  Keep possible zero factors instead of canceling them.
2. Hint 2
  Move both sides together and take out their shared factor.
Answer

$x=-2$ or $x=5$ .

Full solution

Subtract the right side and take out $x+2$ .
$\begin{aligned} (x+2)(x-5)&=0 \end{aligned}$

The Zero Product Property gives $x=-2$ or $x=5$ .
At $-2$ both original sides are zero; at $5$ both are $14$ .

Answer

$x=-2$ or $x=5$ .

Key idea

Moving a shared factor to a zero product preserves the solution where that factor vanishes.
Problem 1: Mark complete
Problem 2 A balance of squares

Solve $(x+1)^2=(2x-1)^2$ for real $x$ .
1. Hint 1
  Move the squares to one side to create a zero expression.
2. Hint 2
  A difference of two squares splits into their difference times their sum.
Answer

$x=0$ or $x=2$ .

Full solution

The difference factors as
$(2-x)(3x)=0$
, so $x=2$ or $x=0$ .
At $0$ , both original squares are $1$ ; at $2$ , both are $9$ .

Answer

$x=0$ or $x=2$ .

Key idea

Equal squares can be compared by factoring their difference.
Problem 2: Mark complete
Problem 3 An unfinished product

The polynomial $f(x)=3x^2-7x-6$ has zero $3$ . Write it as $(x-3)L(x)$ and find the linear expression $L(x)$ .
1. Hint 1
  The leading term of the missing factor must produce the squared coefficient.
2. Hint 2
  Expand $(x-3)(3x+d)$ and match a remaining coefficient.
Answer

$L(x)=3x+2$ .

Full solution

The leading coefficient forces $L(x)=3x+d$ .
Expansion gives $3x^2+(d-9)x-3d$ .
Matching the linear coefficient gives
$d-9=-7$
Hence $d=2$ .
The constant also checks, since $-3d=-6$ .

Answer

$L(x)=3x+2$ .

Key idea

A known zero gives one factor, and coefficient matching recovers the other.
Problem 3: Mark complete
Problem 4 A panel border

A square panel has side length $x$ cm, where $x>0$ . A border adds 1 cm along each edge, making the outer square’s side $x+2$ cm. The outer area is four times the panel area. Find the panel’s side length.
1. Hint 1
  Translate the two areas before moving all terms to one side.
2. Hint 2
  Factor the difference of the two squared expressions, then apply the positive-length condition.
Answer

$x=2$ cm.

Full solution

The area condition is
$(x+2)^2=4x^2.$
Subtract the right side and factor the difference of squares.
$(2-x)(3x+2)=0$
The factors give $x=2$ or $x=-2/3$ .
Only the positive value is a side length.
At $x=2$ cm, the panel area is $4$ square cm and the outer area is $16$ square cm, exactly four times as large.

Answer

$x=2$ cm.

Key idea

A factored area equation can give an algebraic root that the positive-length condition excludes.
Problem 4: Mark complete
Problem 5 A horizontal crossing

Find every intersection of $y=x^2+3x-4$ with the horizontal line $y=6$ . Give each point as an ordered pair.
1. Hint 1
  At an intersection, the two formulas for the height agree.
2. Hint 2
  Move the target height across before factoring.
Answer

$(-5,6)$ and $(2,6)$ .

Full solution

Set the quadratic output equal to $6$ .
$\begin{aligned} x^2+3x-10&=0 \\ (x+5)(x-2)&=0 \end{aligned}$

The roots are $-5$ and $2$ , each at height $6$ .
Checking gives $25-15-4=6$ and $4+6-4=6$ .
These are zeros of the shifted function $f(x)-6$ , not zeros of $f$ itself.

Answer

$(-5,6)$ and $(2,6)$ .

Key idea

A horizontal intersection becomes a zero after the target height is subtracted.
Problem 5: Mark complete
Problem 6 A restricted input range

Find all $x$ in $0\le x\le4$ satisfying $3x(x-2)=x(x+2)$ .
1. Hint 1
  Find all algebraic candidates before applying the input interval.
2. Hint 2
  Bring the products together without dividing by $x$ .
Answer

$x=0$ or $x=4$ .

Full solution

Move the right product to the left.
$\begin{aligned} 2x^2-8x&=0 \\ 2x(x-4)&=0 \end{aligned}$

The roots are $0$ and $4$ , both in the allowed closed interval.
At $0$ both sides vanish; at $4$ both sides equal $24$ .

Answer

$x=0$ or $x=4$ .

Key idea

Solve before applying a domain restriction, and keep zero factors during the algebra.
Problem 6: Mark complete
Problem 7 Recovering a rule

A quadratic has leading coefficient $2$ , constant term $-8$ , and zero $2$ . Find its other zero and its standard form.
1. Hint 1
  Use the known zero to start a factorization.
2. Hint 2
  The leading and constant terms determine the entries of the remaining linear factor.
Answer

Other zero $-2$ ; $f(x)=2x^2-8$ .

Full solution

The known zero gives
$f(x)=(x-2)(2x+d).$
The constant condition is $-2d=-8$ , so $d=4$ .
Thus
$f(x)=2(x-2)(x+2)$
The other zero is $-2$ , and expansion gives $2x^2-8$ .
Both zeros make the rule vanish.

Answer

Other zero $-2$ ; $f(x)=2x^2-8$ .

Key idea

The leading coefficient and constant can finish a factorization started from a known zero.
Problem 7: Mark complete
Problem 8 Building from one zero and an intercept

A quadratic function has $x=1$ as a zero and $y$ -intercept $-6$ . Give two different quadratics, each in factored form with integer coefficients, that satisfy both conditions.
1. Hint 1
  A zero at $1$ gives one linear factor directly.
2. Hint 2
  The $y$ -intercept fixes the product of the leading coefficient and $r$ , the number subtracted in the second factor.
Answer

For example, $(x-1)(x+6)$ and $2(x-1)(x+3)$ (other correct pairs exist).

Full solution

A zero at $x=1$ gives the factor $(x-1)$ , so $f(x)=a(x-1)(x-r)$ for a nonzero constant $a$ and a constant $r$ .
The $y$ -intercept means $f(0)=-6$ .
Since $f(0)=a(-1)(-r)$ , this gives $ar=-6$ .

Any integer factor pair of $-6$ works: $a=1,r=-6$ gives $(x-1)(x+6)$ .
Substituting $x=0$ gives $f(0)=(-1)(6)$ , which is $-6$ .
Likewise $a=2,r=-3$ gives $2(x-1)(x+3)$ ; substituting $x=0$ gives $f(0)=2(-1)(3)$ , which is $-6$ .
Both are valid, and other pairs exist.

Answer

For example, $(x-1)(x+6)$ and $2(x-1)(x+3)$ (other correct pairs exist).

Key idea

One zero and the $y$ -intercept fix a family of quadratics, not a single one, since the leading coefficient and the other zero can trade off.
Problem 8: Mark complete
Problem 9 A limited search

A student tests every integer from $-10$ through $10$ in $x^2=\frac14$ and finds no root. They conclude there are no real roots. Is this justified? Explain and give the real roots.
1. Hint 1
  The list searched contains only integers.
2. Hint 2
  Write the equation as a difference of squares using the square root of the constant.
Answer

No; $x=-\frac12$ or $x=\frac12$ .

Full solution

An integer search says nothing about noninteger candidates.
Here
$(x-\tfrac12)(x+\tfrac12)=0$
so the roots are $1/2$ and $-1/2$ , both real but absent from the searched list.
Each has square $1/4$ .

Answer

No; $x=-\frac12$ or $x=\frac12$ .

Key idea

A failed search excludes its tested candidates, not the rest of the stated domain.
Problem 9: Mark complete
Problem 10 A marked point on a parabola

The figure shows a parabola with an unspecified leading coefficient. Using only what the figure shows, determine whether the factor $(x+2)$ appears once or twice in a factored form for $f$ , and find the sign of the leading coefficient. Explain your reasoning.

The graph of $f$ , with an unspecified leading coefficient.

Text description of this figure

A grid with the horizontal axis running from negative 5 to 1 and the vertical axis running from negative 5 to 1, gridlines and number labels at every whole number. A single smooth curve rises from the lower left, touches the horizontal axis at exactly one marked point, negative 2 comma 0, and then descends back down to the lower right, staying at or below the horizontal axis everywhere. No equation and no other point is shown.
1. Hint 1
  Notice whether the curve crosses to the other side of the axis at the marked point, or stays on the same side.
2. Hint 2
  A repeated factor keeps the same sign on the two sides of its zero.
Answer

$(x+2)$ appears twice; $f(x)=a(x+2)^2$ with $a<0$ .

Full solution

The curve touches the axis at exactly one point, $(-2,0)$ , and does not cross there: it lies on the same side (below the axis) on both sides of the contact point.
A quadratic that touches the axis at its vertex without crossing has a repeated zero there, so
$f(x)=a(x+2)^2$
and the factor $x+2$ occurs twice.
The square is nonnegative everywhere, and the graph stays at or below the axis, so the leading coefficient must be negative: $a<0$ .

Answer

$(x+2)$ appears twice; $f(x)=a(x+2)^2$ with $a<0$ .

Key idea

A repeated zero makes a quadratic touch the axis without changing sign.
Problem 10: Mark complete

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