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Zielflug [23.3K]

2 years ago

14

PLS I NEED HELP ANSWER WILL GIVE BRAINLIEST ANSWER BOTH QUESTIONS ONE AND TWO

1 answer:

KiRa [710]2 years ago

8 0

I hope this is helpful

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I need help with algebra

Arlecino [84]

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Hello, my name is 1frogaroni feel free to ask any questions about algebra such as equations you don't understand. I will try my best to help you out with your question :)

Step-by-step explanation:

Oh and welcome to brainly where the community will be glad to help you!

3 0

2 years ago

2x^3-x^2-3x=210<br> the answer is 5 but I want to know why.

mel-nik [20]

Answer:

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

Step-by-step explanation:

1) Move all terms to one side.

$2x^{3} -x^{2} -3x-210=0$

2) Factor $2{x}^{3}-{x}^{2}-3x-210$ using Polynomial Division.

1 - Factor the following.

$2x^{3} -x^{2} -3x-210$

2 - First, find all factors of the constant term 210.

$1,2,3,4,5,6,7,10,14,15,21,30,35,42,70,105,210$

3) Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

$2*1^{3} -1^{2} -3*1-210=-212$

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

$2(-1)^{3} -(-1)^{2} -3*-1-210=-210$

Substitute 2 into x. Since the result is not 0, x-2 is not a factor..

$2*2^{3} -2^{2} -3*2-210=-204$

Substitute -2 into x. Since the result is not 0, x+2 is not a factor..

$2{(-2)}^{3}-{(-2)}^{2}-3\times -2-210 = -224$

Substitute 3 into x. Since the result is not 0, x-3 is not a factor..

$2\times {3}^{3}-{3}^{2}-3\times 3-210 = -174$

Substitute -3 into x. Since the result is not 0, x+3 is not a factor..

$2{(-3)}^{3}-{(-3)}^{2}-3\times -3-210 = -264$

Substitute 5 into x. Since the result is 0, x-5 is a factor..

$2\times {5}^{3}-{5}^{2}-3\times 5-210 =0$

------------------------------------------------------------------------------------------

⇒ $x-5$

4) Polynomial Division: Divide $2{x}^{3}-{x}^{2}-3x-210$ by $x-5$ .

$2x^{2}$ $9x$ $42$

-------------------------------------------------------------------------

$x-5$ | $2x^{3}$ $-x^{2}$ $-3x$ $-210$

$2x^{3}$ $-10x^{2}$

-----------------------------------------------------------------------

$9x^{2}$ $-3x$ $-210$

--------------------------------------------------------------------------

$42x$ $-210$

$42x$ $-210$

-------------------------------------------------------------------------

5) Rewrite the expression using the above.

$2x^2+9x+42$

$(2x^2+9x+42)(x-5)=0$

3) Solve for $x.$

$x=5$

4) Use the Quadratic Formula.

1 - In general, given $a{x}^{2}+bx+c=0$ , there exists two solutions where:

$x=\frac{-b+\sqrt{b^{2} -4ac} }{2a} ,\frac{-b-\sqrt{b^2-4ac} }{2a}$

2 - In this case, $a=2,b=9$ and $c = 42.$

$x=\frac{-9+\sqrt{9^2*-4*2*42} }{2*2} ,\frac{-9-\sqrt{9^2-4*2*42} }{2*2}$

3 - Simplify.

$x=\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

5) Collect all solutions from the previous steps.

$x=5,\frac{-9+\sqrt{255}i }{4} ,\frac{-9-\sqrt{255}i }{4}$

4 0

2 years ago

Read 2 more answers

Drag the tiles to the boxes to form correct pairs.<br> Match the pairs of equivalent expressions.

Rashid [163]

Answer:

The following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Step-by-step explanation:

Lets solve all the expressions to match the results.

$5\left(2t+1\right)+\left(-7t+28\right)$

<em>Solving the expression</em>

$5\left(2t+1\right)+\left(-7t+28\right)$

$\mathrm{Remove\:parentheses}:\quad \left(-a\right)=-a$

$5\left(2t+1\right)-7t+28$

$10t+5-7t+28$

$3t+33$

Therefore, $5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$

$3\left(3t-4\right)-\left(2t+10\right)$

<em>Solving the expression</em>

$3\left(3t-4\right)-\left(2t+10\right)$

$9t-12-\left(2t+10\right)$

$9t-12-2t-10$

$7t-22$

Therefore, $3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

<em>Solving the expression</em>

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$4t-\frac{8}{5}-\left(3-\frac{4}{3}t\right)$

$4t-\frac{8}{5}-\left(-\frac{4t}{3}+3\right)$

$4t-\frac{8}{5}-3+\frac{4t}{3}$

As

$-3-\frac{8}{5}:\quad -\frac{23}{5}$ and $\frac{4t}{3}+4t:\quad \frac{16t}{3}$

So,

$4t-\frac{8}{5}-3+\frac{4t}{3}$ will become $\frac{16t}{3}-\frac{23}{5}$

Therefore, $\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

<em>Solving the expression</em>

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$

$\mathrm{Remove\:parentheses}:\quad \left(a\right)=a$

$-\frac{9}{2}t+3+\frac{7}{4}t+33$

$\mathrm{Group\:like\:terms}$

$\frac{9}{2}t+\frac{7}{4}t+3+33$

$\mathrm{Add\:similar\:elements:}\:-\frac{9}{2}t+\frac{7}{4}t=-\frac{11}{4}t$

$-\frac{11}{4}t+3+33$

$-\frac{11}{4}t+36$

Therefore, $\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Thus, the following pairs/results are matched:

$5\left(2t+1\right)+\left(-7t+28\right)$ = $3t+33$
$3\left(3t-4\right)-\left(2t+10\right)$ = $7t-22$

$\left(4t-\frac{8}{5}\right)-\left(3-\frac{4}{3}t\right)$ = $\frac{16t}{3}-\frac{23}{5}$

$\left(-\frac{9}{2}t+3\right)+\left(\frac{7}{4}t+33\right)$ = $-\frac{11}{4}t+36$

Keywords: algebraic expression

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