All categories

2 years ago

What is -2(× - 4) in distributive property

1 answer:

6 0

-2x + 8. you multiply all the terms inside the parenthesis by the one outside of/attached to them. the reason it is +8 and not -8 is because the negatives cancel out to give you a positive.

You might be interested in

What is the value of the angle marked with xxx?

Mice21 [21]

I have no clue what to answer 9

5 0

2 years ago

14 POINTS 7x - 3 = 18

Svetlanka [38]

Subtract 3 first so it would be 7x=15
then divide by 7 so x would be approximately 2.14

5 0

3 years ago

Read 2 more answers

According to the 1993 World Almanac, the number of calories a person walking at 3 mph burns per minute depends on the persons we

zaharov [31]

150 pounds is your answer

3 0

3 years ago

Given g(x) = x² - 6x - 16, which statement is true?

inn [45]

The correct choice of this question with the given polynomial is "The zeros are -2 and 8, because the factors of g are (x + 2) and (x - 8)". (Correct choice: H)

<h3>How to analyze a second orden polynomial with constant coefficients</h3>

In this case we have a second order polynomial of the form x² - (r₁ + r₂) · x + r₁ · r₂, whose solution is (x - r₁) · (x - r₂) and where r₁ and r₂ are the roots of the polynomial, which can be real or complex numbers but never both according the fundamental theorem of algebra.

If we know that g(x) = x² - 6 · x - 16, then the factored form of the expression is g(x) = (x - 8) · (x + 2). Hence, the correct choice of this question with the given polynomial is "The zeros are -2 and 8, because the factors of g are (x + 2) and (x - 8)". $\blacksquare$

To learn more on polynomials, we kindly invite to check this verified question: brainly.com/question/11536910

4 0

2 years ago

What is the completely factored form of f(x)=x^3-7x^2+2x+4

xxMikexx [17]

$Solution, \mathrm{Factor}\:x^3-7x^2+2x+4:\quad \left(x-1\right)\left(x^2-6x-4\right)$

$Steps:$

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

$Use\:the\:rational\:root\:theorem,\\a_0=4,\:\quad a_n=1,\\\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:4,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1,\\\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:4}{1},\\\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x-1,\\\left(x-1\right)\frac{x^3-7x^2+2x+4}{x-1}$

$\frac{x^3-7x^2+2x+4}{x-1}$

$\mathrm{Divide}\:\frac{x^3-7x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}x^3-7x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{x^3}{x},\\\mathrm{Quotient}=x^2,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}x^2:\:x^3-x^2,\\\mathrm{Subtract\:}x^3-x^2\mathrm{\:from\:}x^3-7x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-6x^2+2x+4,\\Therefore,\\\frac{x^3-7x^2+2x+4}{x-1}=x^2+\frac{-6x^2+2x+4}{x-1}$

$\mathrm{Divide}\:\frac{-6x^2+2x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-6x^2+2x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-6x^2}{x}=-6x,\\\mathrm{Quotient}=-6x,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-6x:\:-6x^2+6x,\\\mathrm{Subtract\:}-6x^2+6x\mathrm{\:from\:}-6x^2+2x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=-4x+4,\\\mathrm{Therefore},\\\frac{-6x^2+2x+4}{x-1}=-6x+\frac{-4x+4}{x-1}$

$\mathrm{Divide}\:\frac{-4x+4}{x-1},\\\mathrm{Divide\:the\:leading\:coefficients\:of\:the\:numerator\:}-4x+4\mathrm{\:and\:the\:divisor\:}x-1\mathrm{\::\:}\frac{-4x}{x}=-4,\\\mathrm{Quotient}=-4,\\\mathrm{Multiply\:}x-1\mathrm{\:by\:}-4:\:-4x+4,\\\mathrm{Subtract\:}-4x+4\mathrm{\:from\:}-4x+4\mathrm{\:to\:get\:new\:remainder},\\\mathrm{Remainder}=0,\\\mathrm{Therefore},\\\frac{-4x+4}{x-1}=-4$

$\mathrm{The\:Correct\:Answer\:is\:\left(x-1\right)\left(x^2-6x-4\right)}$

$\mathrm{Hope\:This\:Helps!!!}$

$\mathrm{-Austint1414}$

8 0

3 years ago