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3 years ago

Match each polynomial function w/ one of its factors.

Mathematics

2 answers:

gladu [14]3 years ago

8 0

Answer:

See below in bold.

Step-by-step explanation:

The first one looks like x + 3 might be a factor. By the Factor Theorem :To check we plug in x = -3 to see if the f(3) = 0.

f(3) = -27 - 27 + 39 + 15 = 0

So x+3 is a factor of x^3 - 3x^2 - 13x + 15.

x^3 - 2x^2 - x + 2

Because of the last term +2 , it looks like x - 2 might be a factor of this so we evaluate f(2)

f(2) = 8 - 8 - 2 + 2 = 0

So x - 2 is a factor of x^3 - 2x^2 - x + 2 .

x^4 + 3x^3 - 8x^2 + 5x - 25.

Because of the - 25 x + 5 could be a factor so try f(-5)

= (-5)^4 - 3* -125 - 8*25 + 5 * -5 - 25 = 0

So x + 5 is a factor of x^4 + 3x^3 - 8x^2 + 5x - 25.

Finally we check if (x + 4) is a factor of -x^3 + 13x - 12 :

f(-4) = - (-4)^3 + 13*-4 - 12 = 0.

So x + 4 is a factor of -x^3 + 13x - 12.

ololo11 [35]3 years ago

4 0

Answer: 1 goes with 2 4 goes with 3

Step-by-step explanation:

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GIVING BRAINIEST IF CORRECT!!<br><br> Factor x^4 - 16 completely

ioda

Answer:

$(x^2 + 4)(x + 2)(x - 2)$

Step-by-step explanation:

Hello!

We can use the difference of square method.

<h2>Difference Of Squares (DOS)</h2>

The formula for the DOS is $(a + b)(a - b) = a^2 - b^2$

It is a simple way to factor polynomials.

The criteria:

Has to begin and end with a perfect square
The operation has to be subtraction

<h3>Factor:</h3>

$x^4 - 16$

Begins with a perfect square (x² * x²) and ends with a perfect square (4 * 4)

$(x^2 + 4)(x^2 - 4)$

Warning! Watch out, there may be another DOS!

$x^2 - 4$ is another DOS

$(x^2 + 4)(x + 2)(x - 2)$

The x² + 4 is not a DOS because the operation is addition.

The final factored form is $(x^2 + 4)(x + 2)(x - 2)$

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