Solve $$(3x - 8)(x - 1) = 3x$$ by first multiplying the binomials and then writing the quadratic equation in standard form.

**Step 1 — Multiply the binomials:** Expand the left side using FOIL:
$$3x^2 - 11x + 8 = 3x$$

**Step 2 — Write in standard form:** Subtract $$3x$$ from both sides to set the equation to zero:
$$3x^2 - 14x + 8 = 0$$

**Step 3 — Factor the trinomial:**
$$(3x - 2)(x - 4) = 0$$

**Step 4 — Apply the Zero Product Property:** Set each factor to zero:
$$3x - 2 = 0 \quad \text{or} \quad x - 4 = 0$$

**Step 5 — Solve each equation:**
$$3x = 2 \implies x = \frac{2}{3} \quad \text{or} \quad x = 4$$

**Step 6 — Check:** The check is left as an exercise. The solutions are $$x = \frac{2}{3}$$ and $$x = 4$$.

Solving $$(3x - 8)(x - 1) = 3x$$

Solve $$(2m + 1)(m + 3) = 12m$$ by expanding and converting to standard form before factoring.

First, multiply the binomials on the left side: $$2m^2 + 7m + 3 = 12m$$.
Next, subtract $$12m$$ from both sides to write the equation in standard form: $$2m^2 - 5m + 3 = 0$$.
Factor the trinomial: $$(2m - 3)(m - 1) = 0$$.
Apply the Zero Product Property by setting each factor equal to zero: $$2m - 3 = 0$$ or $$m - 1 = 0$$.
Solve the linear equations to find the solutions: $$m = \frac{3}{2}$$ and $$m = 1$$.

Solving $$(2m + 1)(m + 3) = 12m$$

Solve $$(k + 1)(k - 1) = 8$$ by first expanding the left side and moving the constant to the left to reach standard form.

Multiply the binomials (which form a difference of squares): $$k^2 - 1 = 8$$.
Subtract $$8$$ from both sides to write the equation in standard form: $$k^2 - 9 = 0$$.
Factor the resulting difference of squares: $$(k - 3)(k + 3) = 0$$.
Use the Zero Product Property to set each factor to zero: $$k - 3 = 0$$ or $$k + 3 = 0$$.
Solve for the variable to find the solutions: $$k = 3$$ and $$k = -3$$.

Solving $$(k + 1)(k - 1) = 8$$

Solve $$3x^2 = 12x + 63$$ by factoring out a constant greatest common factor first.

**Step 1 — Write in standard form:** Subtract $$12x$$ and $$63$$ from both sides to set the equation to zero:
$$3x^2 - 12x - 63 = 0$$

**Step 2 — Factor the greatest common factor:** The terms share a numerical GCF of $$3$$:
$$3(x^2 - 4x - 21) = 0$$

**Step 3 — Factor the trinomial:**
$$3(x - 7)(x + 3) = 0$$

**Step 4 — Apply the Zero Product Property:** Set each factor equal to zero. Notice that there are three factors, but the first factor is a constant. Since $$3 \neq 0$$, only the variable factors can equal zero:
$$x - 7 = 0 \quad \text{or} \quad x + 3 = 0$$

**Step 5 — Solve each equation:**
$$x = 7 \quad \text{or} \quad x = -3$$

**Step 6 — Check:** The check is left as an exercise. The solutions are $$x = 7$$ and $$x = -3$$.

Solving $$3x^2 = 12x + 63$$

Solve $$18a^2 - 30 = -33a$$ by first writing it in standard form and factoring out a constant.

Add $$33a$$ to both sides and arrange in descending order: $$18a^2 + 33a - 30 = 0$$.
Factor out the greatest common factor of $$3$$: $$3(6a^2 + 11a - 10) = 0$$.
Factor the resulting trinomial: $$3(3a - 2)(2a + 5) = 0$$.
Apply the Zero Product Property. Since the constant factor $$3$$ cannot equal zero ($$3 \neq 0$$), set the variable factors to zero: $$3a - 2 = 0$$ or $$2a + 5 = 0$$.
Solve the linear equations: $$a = \frac{2}{3}$$ and $$a = -\frac{5}{2}$$.

Solving $$18a^2 - 30 = -33a$$

Solve $$123b = -6 - 60b^2$$ by converting the equation to standard form and factoring out a greatest common factor.

Add $$60b^2$$ and $$6$$ to both sides, writing the polynomial in descending order: $$60b^2 + 123b + 6 = 0$$.
Factor out the common factor of $$3$$: $$3(20b^2 + 41b + 2) = 0$$.
Factor the trinomial: $$3(20b + 1)(b + 2) = 0$$.
Using the Zero Product Property, set the variable factors to zero, noting that $$3 \neq 0$$: $$20b + 1 = 0$$ or $$b + 2 = 0$$.
Solve for the variable: $$b = -\frac{1}{20}$$ and $$b = -2$$.

Solving $$123b = -6 - 60b^2$$

Solve the polynomial equation $$8x^3 = 24x^2 - 18x$$ by converting it to standard form and applying factoring techniques.

**Step 1 — Write in standard form.** Subtract $$24x^2$$ and add $$18x$$ to both sides to bring all terms to one side:
$$8x^3 - 24x^2 + 18x = 0$$

**Step 2 — Factor the greatest common factor.** The three terms share a common factor of $$2x$$. Factor it out:
$$2x(4x^2 - 12x + 9) = 0$$

**Step 3 — Factor the trinomial.** The remaining trinomial is a perfect square trinomial:
$$2x(2x - 3)(2x - 3) = 0$$

**Step 4 — Apply the Zero Product Property.** Set each variable factor to zero:
$$2x = 0 \quad \text{or} \quad 2x - 3 = 0 \quad \text{or} \quad 2x - 3 = 0$$

**Step 5 — Solve each equation:**
$$x = 0 \quad \text{or} \quad x = \frac{3}{2} \quad \text{or} \quad x = \frac{3}{2}$$

The distinct solutions are $$x = 0$$ and $$x = \frac{3}{2}$$.

Solving $$8x^3 = 24x^2 - 18x$$ by Factoring

Solve the polynomial equation $$16y^2 = 32y^3 + 2y$$ using factoring and the Zero Product Property.

**Step 1 — Write in standard form.** Subtract $$16y^2$$ from both sides so that one side equals zero. Rearrange in descending order:
$$32y^3 - 16y^2 + 2y = 0$$

**Step 2 — Factor the greatest common factor.** All terms share a common factor of $$2y$$. Factor it out:
$$2y(16y^2 - 8y + 1) = 0$$

**Step 3 — Factor the trinomial.** The resulting trinomial is a perfect square trinomial:
$$2y(4y - 1)(4y - 1) = 0$$

**Step 4 — Apply the Zero Product Property.** Set each variable factor equal to zero:
$$2y = 0 \quad \text{or} \quad 4y - 1 = 0 \quad \text{or} \quad 4y - 1 = 0$$

**Step 5 — Solve each equation:**
$$y = 0 \quad \text{or} \quad y = \frac{1}{4} \quad \text{or} \quad y = \frac{1}{4}$$

The distinct solutions are $$y = 0$$ and $$y = \frac{1}{4}$$.

Solving $$16y^2 = 32y^3 + 2y$$ by Factoring

Solve $$9m^3 + 100m = 60m^2$$ by collecting all terms on one side, factoring out the variable GCF, and then factoring the resulting trinomial.

**Step 1 — Bring all terms to one side.** Subtract $$60m^2$$ from both sides so that one side equals zero:
$$9m^3 - 60m^2 + 100m = 0$$

**Step 2 — Factor the greatest common factor first.** All three terms share a common factor of $$m$$. Factor it out:
$$m(9m^2 - 60m + 100) = 0$$

**Step 3 — Factor the trinomial.** The trinomial $$9m^2 - 60m + 100$$ is a perfect square trinomial. Factor it as:
$$m(3m - 10)(3m - 10) = 0$$

**Step 4 — Apply the Zero Product Property.** Set each factor equal to zero:
$$m = 0 \quad \text{or} \quad 3m - 10 = 0 \quad \text{or} \quad 3m - 10 = 0$$

**Step 5 — Solve each equation:**
$$m = 0 \quad \text{or} \quad m = \frac{10}{3} \quad \text{or} \quad m = \frac{10}{3}$$

**Step 6 — Check your answers.** The check is left to the reader.

The solutions are $$m = 0$$ and $$m = \frac{10}{3}$$. This example demonstrates solving a polynomial equation by factoring out a common variable factor first, and then factoring the remaining quadratic trinomial.

Solving $$9m^3 + 100m = 60m^2$$ by Factoring

Apply the geometry problem-solving strategy to find the dimensions of a rectangular bedroom when its area is $$117$$ square feet and the length is four feet more than the width. First, sketch the rectangle and let $$w$$ represent the width, making the length $$w + 4$$. Using the area formula for a rectangle, $$A = l \cdot w$$, substitute the known values to obtain $$117 = (w + 4)w$$. Distribute the $$w$$ to yield $$117 = w^2 + 4w$$, and subtract $$117$$ from both sides to write the quadratic equation in standard form: $$0 = w^2 + 4w - 117$$. Factor the trinomial into $$(w + 13)(w - 9) = 0$$ and apply the Zero Product Property to find $$w = -13$$ and $$w = 9$$. Because a physical width cannot be negative, discard $$-13$$. The width is $$9$$ feet, and the length is $$9 + 4 = 13$$ feet.

Finding the Dimensions of a Rectangular Bedroom with Area $$117$$ Square Feet

Apply the geometry problem-solving strategy to find the dimensions of a rectangular sign when its area is $$30$$ square feet and the length is one foot more than the width. Let $$w$$ represent the width, making the length $$w + 1$$. Substitute these expressions into the area formula $$A = l \cdot w$$ to obtain $$30 = (w + 1)w$$. Distribute to get $$30 = w^2 + w$$, and subtract $$30$$ from both sides to rewrite the equation in standard form as $$0 = w^2 + w - 30$$. Factor the quadratic expression into $$(w + 6)(w - 5) = 0$$ and apply the Zero Product Property to find $$w = -6$$ or $$w = 5$$. Because a physical dimension must be positive, discard $$-6$$. The width is $$5$$ feet, and the length is $$5 + 1 = 6$$ feet.

Finding the Dimensions of a Rectangular Sign with Area $$30$$ Square Feet

Apply the geometry problem-solving strategy to find the dimensions of a rectangular patio when its area is $$180$$ square feet and the width is three feet less than the length. Let $$l$$ represent the length, making the width $$l - 3$$. Substitute these expressions into the area formula $$A = l \cdot w$$ to get $$180 = l(l - 3)$$. Distribute to produce $$180 = l^2 - 3l$$, and subtract $$180$$ from both sides to rearrange into the standard quadratic form $$0 = l^2 - 3l - 180$$. Factor the trinomial into $$(l - 15)(l + 12) = 0$$. By the Zero Product Property, $$l = 15$$ or $$l = -12$$. Since a physical length cannot be negative, discard $$-12$$. The length is $$15$$ feet, and the width is $$15 - 3 = 12$$ feet.

Finding the Dimensions of a Rectangular Patio with Area $$180$$ Square Feet

To analyze the height of an object thrown upward, modeled by the polynomial function $$h(t) = -16t^2 + 64t + 80$$, we can find its zeros, the time it reaches a specific height, and its height at a specific time. First, to find the zeros which indicate when the object hits the ground, set $$h(t) = 0$$. Factoring the equation $$-16t^2 + 64t + 80 = 0$$ gives $$-16(t^2 - 4t - 5) = 0$$, which factors further to $$-16(t - 5)(t + 1) = 0$$. This yields $$t = 5$$ and $$t = -1$$. Since time cannot be negative, we discard $$t = -1$$, concluding the object hits the ground at 5 seconds. Second, to find when the object is at a specific height of 80 feet, set $$h(t) = 80$$. Subtracting 80 from both sides gives $$-16t^2 + 64t = 0$$. Factoring out the greatest common factor gives $$-16t(t - 4) = 0$$, yielding $$t = 0$$ and $$t = 4$$. Thus, the object is at 80 feet at 0 and 4 seconds. Third, to find the height at a specific time, such as $$t = 2$$ seconds, substitute 2 into the function. The result is $$h(2) = -16(2)^2 + 64(2) + 80 = 144$$. The object is at 144 feet after 2 seconds.

Finding the Zeros and Specific Heights for $$h(t) = -16t^2 + 64t + 80$$

To analyze the height of an object thrown upward, modeled by the polynomial function $$h(t) = -16t^2 + 48t + 160$$, we can find its zeros, the time it reaches a specific height, and its height at a specific time. First, to find the zeros which indicate when the object hits the ground, set $$h(t) = 0$$. Factoring the equation $$-16t^2 + 48t + 160 = 0$$ gives $$-16(t^2 - 3t - 10) = 0$$, which factors further to $$-16(t - 5)(t + 2) = 0$$. This yields $$t = 5$$ and $$t = -2$$. Discarding the negative time, the object hits the ground at 5 seconds. Second, to find when the object is at a specific height of 160 feet, set $$h(t) = 160$$. Subtracting 160 from both sides gives $$-16t^2 + 48t = 0$$. Factoring gives $$-16t(t - 3) = 0$$, yielding $$t = 0$$ and $$t = 3$$. The object is at 160 feet at 0 and 3 seconds. Third, to find the height at a specific time, such as $$t = 1.5$$ seconds, substitute 1.5 into the function. The result is $$h(1.5) = -16(1.5)^2 + 48(1.5) + 160 = 196$$. The object is at 196 feet after 1.5 seconds.

Finding the Zeros and Specific Heights for $$h(t) = -16t^2 + 48t + 160$$

To analyze the height of an object thrown upward, modeled by the polynomial function $$h(t) = -16t^2 + 32t + 128$$, we can find its zeros, the time it reaches a specific height, and its height at a specific time. First, to find the zeros which indicate when the object hits the ground, set $$h(t) = 0$$. Factoring the equation $$-16t^2 + 32t + 128 = 0$$ gives $$-16(t^2 - 2t - 8) = 0$$, which factors further to $$-16(t - 4)(t + 2) = 0$$. This yields $$t = 4$$ and $$t = -2$$. Discarding the negative time, the object hits the ground at 4 seconds. Second, to find when the object is at a specific height of 128 feet, set $$h(t) = 128$$. Subtracting 128 from both sides gives $$-16t^2 + 32t = 0$$. Factoring gives $$-16t(t - 2) = 0$$, yielding $$t = 0$$ and $$t = 2$$. The object is at 128 feet at 0 and 2 seconds. Third, to find the height at a specific time, such as $$t = 1$$ second, substitute 1 into the function. The result is $$h(1) = -16(1)^2 + 32(1) + 128 = 144$$. The object is at 144 feet after 1 second, which is its highest point.

Finding the Zeros and Specific Heights for $$h(t) = -16t^2 + 32t + 128$$

To solve a quadratic equation by factoring, follow a five-step procedure:

**Step 1.** Write the quadratic equation in standard form, $$ax^2 + bx + c = 0$$.
**Step 2.** Factor the quadratic expression on the left side.
**Step 3.** Use the Zero Product Property by setting each factor equal to zero.
**Step 4.** Solve the resulting linear equations.
**Step 5.** Check the solutions by substituting each separately back into the original equation.

Claude

To solve a quadratic equation using the Zero Product Property, follow a three-step procedure:

Step 1. Set each factor equal to zero. The Zero Product Property states that if a product is zero, at least one factor must be zero.

Step 2. Solve the linear equations. Solve each of the resulting simpler equations to find the values of the variable.

Step 3. Check. Substitute each solution separately back into the original equation to verify that both sides produce a true statement.

How to Solve a Quadratic Equation Using the Zero Product Property

https://openstax.org/details/books/intermediate-algebra-2e

OpenStax Intermediate Algebra

How to Solve a Quadratic Equation by Factoring

Solve the quadratic equation $$3c^2 = 10c - 8$$ by applying the factoring method.

**Step 1 — Write in standard form.** Subtract $$10c$$ and add $$8$$ to both sides:
$$3c^2 - 10c + 8 = 0$$

**Step 2 — Factor the quadratic expression.**
$$(3c - 4)(c - 2) = 0$$

**Step 3 — Apply the Zero Product Property.** Set each factor equal to zero:
$$3c - 4 = 0 \quad \text{or} \quad c - 2 = 0$$

**Step 4 — Solve each linear equation:**
$$c = \frac{4}{3} \quad \text{or} \quad c = 2$$

**Step 5 — Check both solutions** by substituting into the original equation $$3c^2 = 10c - 8$$:
For $$c = \frac{4}{3}$$: $$3\left(\frac{4}{3}\right)^2 = 3\left(\frac{16}{9}\right) = \frac{16}{3}$$ and $$10\left(\frac{4}{3}\right) - 8 = \frac{40}{3} - \frac{24}{3} = \frac{16}{3}$$ ✓
For $$c = 2$$: $$3(2)^2 = 12$$ and $$10(2) - 8 = 20 - 8 = 12$$ ✓
The solutions are $$c = \frac{4}{3}$$ and $$c = 2$$.

Solving $$3c^2 = 10c - 8$$ by Factoring

Solve the quadratic equation $$2d^2 - 5d = 3$$ by applying the factoring method.

**Step 1 — Write in standard form.** Subtract $$3$$ from both sides:
$$2d^2 - 5d - 3 = 0$$

**Step 2 — Factor the quadratic expression.**
$$(2d + 1)(d - 3) = 0$$

**Step 3 — Apply the Zero Product Property.** Set each factor equal to zero:
$$2d + 1 = 0 \quad \text{or} \quad d - 3 = 0$$

**Step 4 — Solve each linear equation:**
$$d = -\frac{1}{2} \quad \text{or} \quad d = 3$$

**Step 5 — Check both solutions** by substituting into the original equation $$2d^2 - 5d = 3$$:
For $$d = -\frac{1}{2}$$: $$2\left(-\frac{1}{2}\right)^2 - 5\left(-\frac{1}{2}\right) = 2\left(\frac{1}{4}\right) + \frac{5}{2} = \frac{1}{2} + \frac{5}{2} = 3$$ ✓
For $$d = 3$$: $$2(3)^2 - 5(3) = 18 - 15 = 3$$ ✓
The solutions are $$d = -\frac{1}{2}$$ and $$d = 3$$.

Solving $$2d^2 - 5d = 3$$ by Factoring

Solve the quadratic equation $$169x^2 = 49$$ by applying the factoring method.

**Step 1 — Write in standard form.** Subtract $$49$$ from both sides:
$$169x^2 - 49 = 0$$

**Step 2 — Factor the quadratic expression.** The binomial is a difference of squares.
$$(13x - 7)(13x + 7) = 0$$

**Step 3 — Apply the Zero Product Property.** Set each factor equal to zero:
$$13x - 7 = 0 \quad \text{or} \quad 13x + 7 = 0$$

**Step 4 — Solve each linear equation:**
$$x = \frac{7}{13} \quad \text{or} \quad x = -\frac{7}{13}$$

**Step 5 — Check both solutions** by substituting into the original equation $$169x^2 = 49$$:
For $$x = \frac{7}{13}$$: $$169\left(\frac{7}{13}\right)^2 = 169\left(\frac{49}{169}\right) = 49$$ ✓
For $$x = -\frac{7}{13}$$: $$169\left(-\frac{7}{13}\right)^2 = 169\left(\frac{49}{169}\right) = 49$$ ✓
The solutions are $$x = \frac{7}{13}$$ and $$x = -\frac{7}{13}$$.

Solving $$169x^2 = 49$$ by Factoring

Solve the quadratic equation $$25p^2 = 49$$ by applying the factoring method.

**Step 1 — Write in standard form.** Subtract $$49$$ from both sides:
$$25p^2 - 49 = 0$$

**Step 2 — Factor the quadratic expression.** The binomial is a difference of squares.
$$(5p - 7)(5p + 7) = 0$$

**Step 3 — Apply the Zero Product Property.** Set each factor equal to zero:
$$5p - 7 = 0 \quad \text{or} \quad 5p + 7 = 0$$

**Step 4 — Solve each linear equation:**
$$p = \frac{7}{5} \quad \text{or} \quad p = -\frac{7}{5}$$

**Step 5 — Check both solutions** by substituting into the original equation $$25p^2 = 49$$:
For $$p = \frac{7}{5}$$: $$25\left(\frac{7}{5}\right)^2 = 25\left(\frac{49}{25}\right) = 49$$ ✓
For $$p = -\frac{7}{5}$$: $$25\left(-\frac{7}{5}\right)^2 = 25\left(\frac{49}{25}\right) = 49$$ ✓
The solutions are $$p = \frac{7}{5}$$ and $$p = -\frac{7}{5}$$.

Solving $$25p^2 = 49$$ by Factoring

Solve the quadratic equation $$36x^2 = 121$$ by applying the factoring method.

**Step 1 — Write in standard form.** Subtract $$121$$ from both sides:
$$36x^2 - 121 = 0$$

**Step 2 — Factor the quadratic expression.** The binomial is a difference of squares.
$$(6x - 11)(6x + 11) = 0$$

**Step 3 — Apply the Zero Product Property.** Set each factor equal to zero:
$$6x - 11 = 0 \quad \text{or} \quad 6x + 11 = 0$$

**Step 4 — Solve each linear equation:**
$$x = \frac{11}{6} \quad \text{or} \quad x = -\frac{11}{6}$$

**Step 5 — Check both solutions** by substituting into the original equation $$36x^2 = 121$$:
For $$x = \frac{11}{6}$$: $$36\left(\frac{11}{6}\right)^2 = 36\left(\frac{121}{36}\right) = 121$$ ✓
For $$x = -\frac{11}{6}$$: $$36\left(-\frac{11}{6}\right)^2 = 36\left(\frac{121}{36}\right) = 121$$ ✓
The solutions are $$x = \frac{11}{6}$$ and $$x = -\frac{11}{6}$$.

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