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3 years ago

X^3+13x^2+32x+20 please help me factorizing above equation

1 answer:

7 0

$x^3+13x^2+32x+20=x^3+x^2+12x^2+ 12x+20x+20 =\\\\=x^2(x+1)+12x(x + 1 )+20(x+1 )=(x+1)(x^2+12x+20)=\\\\=(x+1)[(x^2+10x+2x+20)]=(x+1)[x(x +10)+2(x+10))]=\\\\=(x+1) (x +2) (x+10)$

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3 0

3 years ago

I'll give you brainlies if you help me on this!

ryzh [129]

Answer:

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3 0

2 years ago

Read 2 more answers

What is 16 tens + 4 ones

aalyn [17]

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3 years ago

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Find the length of the following curve. If you have a grapher, you may want to graph the curve to see what it looks like.

stepladder [879]

The length of the curve $y = \frac{1}{27}(9x^2 + 6)^\frac 32$ from x = 3 to x = 6 is 192 units

<h3>How to determine the length of the curve?</h3>

The curve is given as:

$y = \frac{1}{27}(9x^2 + 6)^\frac 32$ from x = 3 to x = 6

Start by differentiating the curve function

$y' = \frac 32 * \frac{1}{27}(9x^2 + 6)^\frac 12 * 18x$

Evaluate

$y' = x(9x^2 + 6)^\frac 12$

The length of the curve is calculated using:

$L =\int\limits^a_b {\sqrt{1 + y'^2}} \, dx$

This gives

$L =\int\limits^6_3 {\sqrt{1 + [x(9x^2 + 6)^\frac 12]^2}\ dx$

Expand

$L =\int\limits^6_3 {\sqrt{1 + x^2(9x^2 + 6)}\ dx$

This gives

$L =\int\limits^6_3 {\sqrt{9x^4 + 6x^2 + 1}\ dx$

Express as a perfect square

$L =\int\limits^6_3 {\sqrt{(3x^2 + 1)^2}\ dx$

Evaluate the exponent

$L =\int\limits^6_3 {3x^2 + 1} \ dx$

Differentiate

$L = x^3 + x|\limits^6_3$

Expand

L = (6³ + 6) - (3³ + 3)

Evaluate

L = 192

Hence, the length of the curve is 192 units