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4 years ago

BIG IDEAS MATH 4.4 Factor y3−5y2+6y−30 completely

Mathematics

1 answer:

azamat4 years ago

8 0

Answer:

(y -5)(y -i√6)(y +i√6)

Step-by-step explanation:

The expression can be factored by grouping:

(y^3 -5y^2) +(6y -30) = y^2(y -5) +6(y -5) = (y^2 +6)(y -5)

The quadratic factor is the difference of squares, if you consider -6 to be a square. In that view, it can be factored as ...

y^2 +6 = (y -i√6)(y +i√6)

Then the complete factorization is ...

(y -5)(y -i√6)(y +i√6)

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Read 2 more answers

WILL MARK BRAINLIEST!!! 40 POINTS!! ACTUAL ANSWERS, PLZZZ

steposvetlana [31]

Answer:

Part A:

$\left(x + 7\right)^{5}=x^{5} + 35 x^{4} + 490 x^{3} + 3430 x^{2} + 12005 x + 16807$

Part B:

The closure property describes cases when mathematical operations are CLOSED. It means that if you apply certain mathematical operations in a polynomial it will still be a polynomial. Polynomials are closed for sum, subtraction, and multiplication.

It means:

$\text{Sum of polynomials } \Rightarrow \text{ It will always be a polynomial}$

$\text{Subtraction of polynomials } \Rightarrow \text{ It will always be a polynomial}$

$\text{Multiplication of polynomials } \Rightarrow \text{ It will always be a polynomial}$

But when it is about division:

$\text{Division of polynomials } \Rightarrow \text{ It will not always/sometimes be a polynomial}$

Example of subtraction of polynomials:

$(2x^2+2x+3) - (x^2+5x+2)$

$x^2-3x+1$

Step-by-step explanation:

First, it is very important to define what is a polynomial in standard form:

It is when the terms are ordered from the highest degree to the lowest degree.

Therefore I can give:

$x^5-5x^4+3x^3-3x^2+7x+20$

but,

$x^5+3x^3-3x^2+7x+20-5x^4$ is not in standard form.

For this question, I can simply give the answer: $x^5-5x^4+3x^3-3x^2+7x+20$ and it is correct.

But I will create a fifth-degree polynomial using this formula

$(a+b)^n=\sum_{k=0}^n \binom{n}{k} a^{n-k} b^k$

Also, note that

$\binom{n}{k}=\frac{n!}{(n-k)!k!}$

For $a=x \text{ and } b=7$

$\left(x + 7\right)^{5}=\sum_{k=0}^{5} \binom{5}{k} \left(x\right)^{5-k} \left(7\right)^k$

$\text{Solving for } k \text{ values: } 0, 1, 2, 3, 4 \text{ and } 5$

Sorry but I will not type every step for each value of $k$

The first one is enough.

For $k=0$

$\binom{5}{0} \left(x\right)^{5-0} \left(7\right)^{0}=\frac{5!}{(5-0)! 0!}\left(x\right)^{5} \left(7\right)^{0}=\frac{5!}{5!} \cdot x^5= x^{5}$

Doing that for $k$ values:

$\left(x + 7\right)^{5}=x^{5} + 35 x^{4} + 490 x^{3} + 3430 x^{2} + 12005 x + 16807$

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3 years ago