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3 years ago

14

I am thinking of a number (s).

2 answers:

Irina-Kira [14]3 years ago

6 0

(?) X 4 - 18 = 94
28 X 4 - 18 = 94

attashe74 [19]3 years ago

3 0

Step-by-step explanation:

?×4-18=94

(28)×4-18=94 I'm just trying

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How does the word equidistant relate to the term perpendicular bisector?

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Ex, If point X was 5 ft away from one endpoint, it would be 5 feet away from the other as well.

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Diego examines his spending and discovers that when he took his new job, he also started buying

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He will save 24 dollars.

Step-by-step explanation:

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2.7 (7.8)??? I don't know how to work this problem...

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Select all of the integers less than -3<br><br>1: -12<br>2: 0<br>3: -4<br>4: 3<br>5: -10

MatroZZZ [7]

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3 years ago

Help appreciated on question in image!<br> Thanks:)

Verdich [7]

Answer:

$x=-1,\:x=-7,\:x=i,\:x=-i$

Step-by-step explanation:

Considering the equation

$x^4+8x^3+8x^2+8x+7=0$

Solving

$x^4+8x^3+8x^2+8x+7$

$\mathrm{Factor\:}x^4+8x^3+8x^2+8x+7:\quad \left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

As

$\mathrm{Use\:the\:rational\:root\:theorem}$

$a_0=7,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:7,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:7}{1}$

$-\frac{1}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+1$

$=\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

Solving

$\frac{x^4+8x^3+8x^2+8x+7}{x+1}$

$=x^3+7x^2+x+7$

Putting $\frac{x^4+8x^3+8x^2+8x+7}{x+1}$ = $x^3+7x^2+x+7$ in equation [A]

So,

$\left(x+1\right)\frac{x^4+8x^3+8x^2+8x+7}{x+1}...[A]$

$=\left(x+1\right)x^3+7x^2+x+7$

As

$x^3+7x^2+x+7=\left(x+7\right)\left(x^2+1\right)$

So,

Equation [A] becomes

$=\left(x+1\right)\left(x+7\right)\left(x^2+1\right)$

So, the polynomial equation becomes

$\left(x+1\right)\left(x+7\right)\left(x^2+1\right)=0$

$\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)$ $\mathrm{Solve\:}\:x+1=0:\quad x=-1$

$\mathrm{Solve\:}\:x+7=0:\quad x=-7$

$\mathrm{Solve\:}\:x^2+1=0:\quad x=i,\:x=-i$

$\mathrm{The\:solutions\:are}$

$x=-1,\:x=-7,\:x=i,\:x=-i$

Keywords: polynomial equation

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3 years ago

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