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 $$k = -3$$.

Claude

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.

How to Solve a Quadratic Equation by Factoring

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

OpenStax Intermediate Algebra

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

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.

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