Resolve 4 theta - 3 theta + theta - 1 into factors.

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To resolve the expression tan⁴ θ − tan³ θ + tan θ − 1 into factors. Step 1: Let x = tan θ. Substitute x into the expression. x⁴ − x³ + x − 1 Step 2: Group the terms. (x⁴ − x³) + (x − 1) Step 3: Factor out the common term from the first group. x³(x − 1) + (x − 1) Step 4: Factor out the common binomial factor (x − 1). (x − 1)(x³ + 1) Step 5: Recognize that x³ + 1 is a sum of cubes, which can be factored using the formula a³ + b³ = (a+b)(a² − ab + b²). Here, a=x and b=1. (x − 1)(x + 1)(x² − x + 1) Step 6: Substitute back tan θ for x. (tan θ − 1)(tan θ + 1)(tan² θ − tan θ + 1) The factored expression is: (tan θ − 1)(tan θ + 1)(tan² θ − tan θ + 1)

Resolve 4 theta - 3 theta + theta - 1 into factors.

Step-by-step solution

Let $x = \tan \theta$ .

Group the terms.

Factor out the common term from the first group.

Factor out the common binomial factor $(x - 1)$ .

Recognize that $x^3 + 1$ is a sum of cubes, which can be factored using the formula $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ .

Substitute back $\tan \theta$ for $x$ .

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More Mathematics Questions

Resolve 4 theta - 3 theta + theta - 1 into factors.

Step-by-step solution

Let x=tan⁡θx = \tan \thetax=tanθ.

Group the terms.

Factor out the common term from the first group.

Factor out the common binomial factor (x−1)(x - 1)(x−1).

Recognize that x3+1x^3 + 1x3+1 is a sum of cubes, which can be factored using the formula a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a+b)(a^2 - ab + b^2)a3+b3=(a+b)(a2−ab+b2).

Substitute back tan⁡θ\tan \thetatanθ for xxx.

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More Mathematics Questions

Let $x = \tan \theta$ .

Factor out the common binomial factor $(x - 1)$ .

Recognize that $x^3 + 1$ is a sum of cubes, which can be factored using the formula $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$ .

Substitute back $\tan \theta$ for $x$ .