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3 years ago

Help me with my math please??

Mathematics

2 answers:

alisha [4.7K]3 years ago

8 0

Put in the (-3) as the x for the problem. PEMDAS teaches us to do multiplication first so you do (-3)*(-5). You get (+15). 15+8=23. 23 is your answer :)

NNADVOKAT [17]3 years ago

8 0

23 is your answer
-5x-3+8

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2 years ago

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rusak2 [61]

Answer:

The zeros are

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Step-by-step explanation:

We have been given the equation x^4-6x^2-7x-6=0

Use rational root theorem, we have

$a_0=6,\:\quad a_n=1$

$\mathrm{The\:dividers\:of\:}a_0:\quad 1,\:2,\:3,\:6,\:\quad \mathrm{The\:dividers\:of\:}a_n:\quad 1$

$\mathrm{Therefore,\:check\:the\:following\:rational\:numbers:\quad }\pm \frac{1,\:2,\:3,\:6}{1}$

$-\frac{2}{1}\mathrm{\:is\:a\:root\:of\:the\:expression,\:so\:factor\:out\:}x+2$

$=\left(x+2\right)\frac{x^4-6x^2-7x-6}{x+2}\\$

$=x^3-2x^2-2x-3$

Again factor using the rational root test, we get

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

Using the zero product rule, we have

$x+2=0:\quad x=-2\\x-3=0:\quad x=3\\x^2+x+1=0\\\\x_{1,\:2}=\frac{-1\pm \sqrt{1^2-4\cdot \:1\cdot \:1}}{2\cdot \:1}\\\\=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

Therefore, the zeros are

$x=-2,\:x=3,\:x=-\frac{1}{2}+i\frac{\sqrt{3}}{2},\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$

4 0

2 years ago

Read 2 more answers

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stepladder [879]

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2 years ago

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4vir4ik [10]

Answer:

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Step-by-step explanation:

First solve parenthesis,

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(12) -154

Finally subtract,

12-154

= -142

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2 years ago

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Ne4ueva [31]

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3 0

2 years ago