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3 years ago

HELP DISTRIBUTIVE PROPERTY QUADRATICS PLZZZZZZZ

Mathematics

1 answer:

Vlad1618 [11]3 years ago

4 0

1: x² + 10x + 25
2: x² - x -12
3: -2x² + 10x
4: -2x² + 9x + 5
5: n² - 4
---------------------------
1: x² + 14x + 49
2: x² - 14x + 49
3: 4n² - 20n + 25
4: 4n² + 20n +25
5: you just have to switch the signs

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Find the quotient. Justify your answer. x^5+2x^4-7x^2-19x+15/x^2+2x+5

vichka [17]

Answer:

$\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5} = x^3-5x+3$

Step-by-step explanation:

To find the quotient we need to apply long division on $\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}$ , as follows:

Divide $\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}$

Divide the leading coefficients of the numerator $x^5+2x^4-7x^2-19x+15$ and the divisor $x^2+2x+5$

$\frac{x^5}{x^2}=x^3$

Quotient = $x^3$

Multiply $x^2+2x+5$ by $x^3$ = $x^5+2x^4+5x^3$

Subtract $x^5+2x^4+5x^3$ from $x^5+2x^4-7x^2-19x+15$ to get new remainder

Remainder = $-5x^3-7x^2-19x+15$

Therefore

$\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}=x^3+\frac{-5x^3-7x^2-19x+15}{x^2+2x+5}$

2. Divide $\frac{-5x^3-7x^2-19x+15}{x^2+2x+5}$

Divide the leading coefficients of the numerator $-5x^3-7x^2-19x+15$ and the divisor $x^2+2x+5$

$\frac{-5x^3}{x^2}=-5x$

Quotient = $-5x$

Multiply $x^2+2x+5$ by $-5x$ = $-5x^3-10x^2-25x$

Subtract $-5x^3-10x^2-25x$ from $-5x^3-7x^2-19x+15$ to get new remainder

Remainder = $3x^2+6x+15$

Therefore

$\frac{-5x^3-7x^2-19x+15}{x^2+2x+5}=-5x+\frac{3x^2+6x+15}{x^2+2x+5}$

$\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5}=x^3-5x+\frac{3x^2+6x+15}{x^2+2x+5}$

3. Divide $\frac{3x^2+6x+15}{x^2+2x+5}$

Divide the leading coefficients of the numerator $3x^2+6x+15$ and the divisor $x^2+2x+5$

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

Quotient = 3

Multiply $x^2+2x+5$ by 3 = $\:3x^2+6x+15$

Subtract $\:3x^2+6x+15$ from $3x^2+6x+15$ to get new remainder

Remainder = 0

Therefore

$\frac{3x^2+6x+15}{x^2+2x+5}=3$

$\frac{x^5+2x^4-7x^2-19x+15}{x^2+2x+5} = x^3-5x+3$

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