Question

math1111: college algebra(11446) lesson 14 - solving polynomial equations by factoring lesson 14 - solving polynomial equations by factoring score: 72/100 answered: 14/19 question 15 find all solutions for the equation below. all answers should be exact, which means no decimals will be accepted. 15c = 81c³ - 126c² c = all answers must be fully simplified. if any numbers can be combined, combine them. any unsimplified fractions or radicals will not receive credit! question help: video message instructor submit question

Explanation:

Step1: Rearrange the equation

Move all terms to one - side to set the equation equal to zero.
$81c^{3}-126c^{2}-15c = 0$

Step2: Factor out the greatest common factor

The GCF of $81c^{3}$, $-126c^{2}$ and $-15c$ is $3c$.
$3c(27c^{2}-42c - 5)=0$

Step3: Factor the quadratic expression

Factor $27c^{2}-42c - 5$. We need to find two numbers that multiply to $27\times(-5)=-135$ and add up to $-42$. The numbers are $-45$ and $3$.
$27c^{2}-45c+3c - 5 = 0$
$9c(3c - 5)+1(3c - 5)=0$
$(3c - 5)(9c + 1)=0$

Step4: Set each factor equal to zero

Set $3c=0$, $3c - 5=0$ and $9c+1 = 0$.
For $3c=0$, $c = 0$.
For $3c - 5=0$, $3c=5$, $c=\frac{5}{3}$.
For $9c+1 = 0$, $9c=-1$, $c=-\frac{1}{9}$.

Answer:

$c = 0,c=\frac{5}{3},c=-\frac{1}{9}$