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Math  /  Algebra

QuestionFind the error in the calculations below, if there is one: Line (1): −3y3+9y2+30y-3 y^{3}+9 y^{2}+30 y Line (2): =−3y(y2−3y−10)=-3 y\left(y^{2}-3 y-10\right) Line (3): =−3y(y2−5y+2y−10)=-3 y\left(y^{2}-5 y+2 y-10\right) Line (4): =−3y(y(y−5)+2(y−5))=-3 y(y(y-5)+2(y-5)) Line (5): =−3y(y+2)(y−5)=-3 y(y+2)(y-5) There are no errors. The error occurred from line (1) to line (2). The error occurred from line (3) to line (4). The error occurred from line (2) to line (3).

Studdy Solution

STEP 1

What is this asking? We're playing detective and hunting for a mistake in some algebra, specifically in factoring a polynomial! Watch out! Factoring can be tricky, so keep an eye on those signs and make sure the distributive property is being used correctly at every step.

STEP 2

1. Check the first factoring step.
2. Verify the splitting of the middle term.
3. Examine the factoring by grouping.

STEP 3

Let's **carefully** examine the transition from Line (1) to Line (2).
We start with −3y3+9y2+30y-3y^3 + 9y^2 + 30y.
We see that −3y-3y is factored out.
Let's distribute that −3y-3y back in to see if we get the original expression: −3y⋅y2=−3y3-3y \cdot y^2 = -3y^3, −3y⋅(−3y)=9y2-3y \cdot (-3y) = 9y^2, and −3y⋅(−10)=30y-3y \cdot (-10) = 30y.
Adding those together gives us −3y3+9y2+30y-3y^3 + 9y^2 + 30y, which **perfectly matches** Line (1)!
So, no errors here.

STEP 4

Now, let's look at how the middle term, −3y-3y, is split in Line (3).
We have −5y+2y-5y + 2y, which indeed adds up to −3y-3y.
Also, notice that (−5)⋅2=−10(-5) \cdot 2 = -10, which is the **same** as the constant term.
This splitting is done to set up factoring by grouping, and it's done correctly!

STEP 5

Finally, let's **investigate** the transition from Line (3) to Line (4).
We have y2−5y+2y−10y^2 - 5y + 2y - 10.
Factoring by grouping, we get y(y−5)+2(y−5)y(y-5) + 2(y-5).
This is **exactly** what Line (4) shows!
Then, factoring out the common factor (y−5)(y-5) gives us (y+2)(y−5)(y+2)(y-5), which, when multiplied by the −3y-3y from earlier steps, gives us Line (5): −3y(y+2)(y−5)-3y(y+2)(y-5).
No errors here either.

STEP 6

There are no errors in the provided calculations.
The factoring was performed correctly at each step!

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