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2 years ago

Help algebra 2 please

Mathematics

2 answers:

Marina CMI [18]2 years ago

7 0

Answer:

x^4 - x^2 - 2x - 1

Step-by-step explanation:

(x^2+x+1)(x^2-x-1) = x^4 - x^3 - x^2 + x^3 - x^2 - x + x^2 - x - 1

= x^4 - x^2 - 2x - 1

34kurt2 years ago

4 0

Answer:

option 4

Step-by-step explanation:

(f*g)(x) =(x² + x+ 1)*(x² - x -1)

= x²*(x² - x -1) + x(x² - x -1) + 1*( x² - x -1)

= x²*x² - x²*x -x²*1 + x*x² - x*x -x*1 + x² - x -1

= x⁴ - x³ - x² +x³ - x² - x + x² - x -1

= x⁴ - x³ + x³ - x² - x² + x² - x - x - 1

= x⁴ - x² - 2x - 1

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3 years ago

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3 years ago

Read 2 more answers

What is the completely factored form of d4 -8d2 + 16

irinina [24]

According to Vieta's Formulas, if $x_1,x_2$ are solutions of a given quadratic equation:

$ax^2+bx+c=0$

Then:

$a(x-x_1)(x-x_2)$ is the completely factored form of $ax^2+bx+c$ .

If choose $x=d^2$ , then:

$\displaystyle x^2-8x+16=0\\\\x_{1,2}= \frac{8\pm \sqrt{64-64} }{2}=4$

So, according to Vieta's formula, we can get:

$x^2-8x+16=(x-4)(x-4)= (x-4)^2$

But $x=d^2$ :

$d^4-8d^2+16=(d^2-4)^2=[(d+2)(d-2)]^2=(d+2)^2(d-2)^2$

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3 years ago