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3 years ago

Find (x³+x²-20x+24) divided by (x-3)

Mathematics

1 answer:

Rus_ich [418]3 years ago

3 0

Answer:

$\frac{x^3 + x^2 -20x + 24}{x-3} = x^2 +4x-8$

Step-by-step explanation:

Given;

p(x) = x³ + x²- 20x + 24

factor of p(x) = x-3

Apply long division method;

x² + 4x - 8

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

x - 3 √ x³ + x²- 20x + 24

-( x³ - 3x²)

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

4x² - 20x + 24

-(4x² -12x )

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

-8x + 24

- (-8x + 24)

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

$thus, \ \frac{x^3 + x^2 -20x + 24}{x-3} = x^2 +4x-8$

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Simplify : x ^ (1/3) * (x ^ (1/2) + 2x ^ 2)

Rudiy27

Answer:

$x^{\frac{1}{3}}\:\times \left(x^{\frac{1}{2}}\:+\:2x^2\right)=x^{\frac{5}{6}}+2x^{\frac{7}{3}}$

Step-by-step explanation:

$x^{\frac{1}{3}}\:\times \left(x^{\frac{1}{2}}\:+\:2x^2\right)$

Let us simplify the expression

$x^{\frac{1}{3}}\:\times \left(x^{\frac{1}{2}}\:+\:2x^2\right)$

$\mathrm{Apply\:the\:distributive\:law}:\quad \:a\left(b+c\right)=ab+ac$

$a=x^{\frac{1}{3}},\:b=x^{\frac{1}{2}},\:c=2x^2$

$=x^{\frac{1}{3}}x^{\frac{1}{2}}+x^{\frac{1}{3}}\times \:2x^2$

$=x^{\frac{1}{3}}x^{\frac{1}{2}}+2x^2x^{\frac{1}{3}}$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^b\cdot \:a^c=a^{b+c}$

so the expression becomes

$=x^{\frac{5}{6}}+2x^{\frac{7}{3}}$

Thus,

$x^{\frac{1}{3}}\:\times \left(x^{\frac{1}{2}}\:+\:2x^2\right)=x^{\frac{5}{6}}+2x^{\frac{7}{3}}$

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3 years ago

I’m overthinking it ( special right triangles - pt )

Black_prince [1.1K]

Answer:

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2 years ago

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4 years ago

The equation w = 250 + 132.5t what are 2 solutions to the equation

yulyashka [42]

Answer:

Step-by-step explanation:

Given:

w = 250 + 132.5t

The two possible solution to the equation includes

1. If the value of w is given, the w value will be substituted in the equation to find the value of t

2. If the value of t is known , the value of t will be substituted in the equation to find the value of w

That is,

w = 250 + 132.5t

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2 years ago

Find (x³+x²-20x+24) divided by (x-3)​

Find (x³+x²-20x+24) divided by (x-3)