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Question Solve the equation 3z(5z−3)=−5z+33 z(5 z-3)=-5 z+3 by factoring. Answer: z={0,35}z=\{0, \frac{3}{5}\}.

Studdy Solution

STEP 1

Assumptions
1. We are given the equation 3z(5z−3)=−5z+33z(5z - 3) = -5z + 3.
2. We will solve the equation by factoring.
3. The solutions will be the values of zz that satisfy the equation.

STEP 2

First, we need to bring all terms to one side of the equation to set it equal to zero.
3z(5z−3)+5z−3=03z(5z - 3) + 5z - 3 = 0

STEP 3

Now, distribute the 3z3z across the terms inside the parentheses.
15z2−9z+5z−3=015z^2 - 9z + 5z - 3 = 0

STEP 4

Combine like terms.
15z2−4z−3=015z^2 - 4z - 3 = 0

STEP 5

Next, we will factor the quadratic equation. We are looking for two numbers that multiply to 15×(−3)=−4515 \times (-3) = -45 and add up to −4-4.

STEP 6

The numbers that satisfy these conditions are −9-9 and 55 because −9×5=−45-9 \times 5 = -45 and −9+5=−4-9 + 5 = -4.

STEP 7

Rewrite the middle term of the quadratic equation using the numbers found in STEP_6.
15z2−9z+5z−3=015z^2 - 9z + 5z - 3 = 0

STEP 8

Now, factor by grouping. Group the first two terms together and the last two terms together.
(15z2−9z)+(5z−3)=0(15z^2 - 9z) + (5z - 3) = 0

STEP 9

Factor out the greatest common factor from each group.
3z(5z−3)+1(5z−3)=03z(5z - 3) + 1(5z - 3) = 0

STEP 10

Notice that we have a common factor of (5z−3)(5z - 3) in both terms.
(3z+1)(5z−3)=0(3z + 1)(5z - 3) = 0

STEP 11

Now, apply the zero-product property, which states that if a product of two factors is zero, then at least one of the factors must be zero.
3z+1=0or5z−3=03z + 1 = 0 \quad \text{or} \quad 5z - 3 = 0

STEP 12

Solve the first equation for zz.
3z+1=03z + 1 = 0 3z=−13z = -1 z=−13z = -\frac{1}{3}

STEP 13

Solve the second equation for zz.
5z−3=05z - 3 = 0 5z=35z = 3 z=35z = \frac{3}{5}

STEP 14

We have found two solutions for zz.
z=−13orz=35z = -\frac{1}{3} \quad \text{or} \quad z = \frac{3}{5}
Answer: z=−13,35z = -\frac{1}{3}, \frac{3}{5}

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