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3 years ago

Which polynomial is factored completely?

Mathematics

2 answers:

Brums [2.3K]3 years ago

5 0

(3x)(9x^2+1) is factored completely since you cannot factor a sum of squares.

oksian1 [2.3K]3 years ago

3 0

Answer:

Option C. is the correct option.

Step-by-step explanation:

We will try to factorize each option

a) $4(4x^{4}-1)$

= 4[(2x²)² - 1] = 4[(2x²-1)(2x²+1)] [since (a²- b²)=(a-b)(a+b)]

b) 2x(y³-4y²+5y)

= 2xy(y² - 4y + 5)

c) 3x(9x²+ 1) It is in the factored form.We can not factorize it further.

d) 5x²- 17x + 14

= 5x²- 10x - 7x + 14 = 5x(x - 2) - 7(x -2) = (5x - 7)(x - 2)

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Find x in This equation.<br> 2x+2=-4(x-5)

san4es73 [151]

Hey there!

2x + 2 = -4(x - 5)

2x + 2 = -4(x) - 4(-5)

2x + 2 = -4x + 20

ADD 4x to BOTH SIDES

2x + 2 + 4x = -4x + 20 + 4x

SIMPLIFY IT!

6x + 2 = 20

SUBTRACT 18 to BOTH SIDES

6x + 2 - 2 = 20 - 2

CANCEL out: 2 - 2 because it give you 0

KEEP: 20 - 2 because it help solve for the x-value

NEW EQUATION: 6x = 20 - 2

SIMPLIFY IT!

6x = 18

DIVIDE 6 to BOTH SIDES

6x/6 = 18/6

CANCEL out: 6/6 because it give you 1

KEE: 18/6 because it help solve for the x-value

NEW EQUATION: x = 18/6

SIMPLIFY IT!

x = 3

Therefore, your answer is: x = 3

Good luck on your assignment and enjoy your day!

~Amphitrite1040:)

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3 years ago

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alex41 [277]

Answer:

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3 years ago

Find the equation of the tangent line to the curve y equals left parenthesis x minus 3 right parenthesis superscript 7 baseline

Rashid [163]

We have been given the expression to be $y=(x-3)^7(x+1)^2$

Since we need to find the tangent at a point, we will have to find the derivative of $y$ as the slope of the tangent at a given point on the curve is always equal to value of the derivative at that point.

Thus, we have to find $\frac{dy}{dx}=\frac{d}{dx} (x-3)^7(x+1)^2$

We will use the product rule of derivatives to find $\frac{dy}{dx}$

Thus, $y'=7(x-3)^6(x+1)^2+(x-3)^7\times2(x+1)$ (using the product rule which states that $(fg)'=f'g+fg'$ )

Taking the common factors out we get:

$y'=(x-3)^6(x+1)(7(x+1)+2(x-3))$

$y'=(x-3)^6(x+1)(7x+7+2x-6)=(x-3)^6(x+1)(9x+1)$

Thus, $y'$ at $x=4$ is given by:

$y'_{x=4}$ =Slope of the tangent of y at x=4= $m$

Thus, $m=(4-3)^6(4+1)(9\times4+1)=185$

Now, the equation of the tangent line which passes through $(x_{1}, y_{1})$ and has slope m is given by:

$y-y_{1}=m(x-x_{1})$

Thus, the equation of the tangent line which passes through $(4,25)$ and has the slope 185 is $y-25=185(x-4)$

Which can be simplified to $y=185x-740+25=185x-715$

Thus, $y=185x-715$

This is the required equation of the tangent.

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3 years ago