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2 years ago

Find the product of the binomials, (x - 3)(x + 2) (x + 4) =

Mathematics

2 answers:

MrRa [10]2 years ago

7 0

Answer:

x^3+3x^2-10x-24

Step-by-step explanation:

(x - 3)(x + 2) (x + 4)

multiply the first two together first

(x^2+2x-3x-6)(x+4) add like terms first

(x^2-x-6)(x+4)

then multiply

x^3+4x^2-x^2-4x-6x-24

add like terms

x^3+3x^2-10x-24

erastovalidia [21]2 years ago

6 0

Answer:

Step-by-step explanation:

(x - 3)(x + 2) (x + 4) =

x^3+3x^2-10x-24

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2 years ago

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$k:6x-14y=-28\ \ \ \ |subtract\ 6x\ from\ both\ sides\\\\-14y=-6x-28\ \ \ \ \ \ |divide\ both\ sides\ by\ (-14)\\\\y=\frac{-6}{-14}x-\frac{28}{-14}\\\\y=\frac{3}{7}x+2\\----------------------------\\l:3y-7x=-14\ \ \ \ \ |add\ 7x\ to\ both\ sides\\\\3y=7x-14\ \ \ \ \ \ |divide\ both\ sides\ by\ 3\\\\y=\frac{7}{3}x-\frac{14}{3}$

$Two\ lines\ are\ perpendicular\ if\ product\ of\ the\ slopes\ is\ equal\ -1.\\\\k:y=\frac{3}{7}x+2\to the\ skolpe\ m_k=\frac{3}{7}\\\\l:y=\frac{7}{3}x-\frac{14}{3}\to the\ slope\ m_l=\frac{7}{3}\\\\m_k\times m_l=\frac{3}{7}\times\frac{7}{3}=1\neq-1\\\\conclusion:the\ lines\ are\ not\ perpendicular\\\\Two\ lines\ are\ parallel\ if\ the\ slopes\ are\ equal.\\\\m_k=\frac{3}{7};\ m_l=\frac{7}{3}\to m_k\neq m_l\\\\conclusion:the\ lines\ are\ not\ parallel$

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