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3 years ago

a)postos pelo vértice b )adjacentes suplementares c ) congruentes e não opostos pelo vértice d )suplementares e não adjacentes

Mathematics

1 answer:

Tpy6a [65]3 years ago

7 0

I can not read this because it is a different language sorry

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Let f(x) = -3x2 + 6x. Find f(2)

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Write each expression as a product of linear factors 1. X^2+1/2x 2. X^2+2x-3 3. (2x-3)^2 4. X^3+2x^2-19x-20

Grace [21]

Answer:

1. X^2+1/2x = $x(x+\frac{1}{2})$

2. X^2+2x-3 = $(x-1)(x+3)$

3. (2x-3)^2 = (2x-3)(2x-3)

4. X^3+2x^2-19x-20 = $(x+1)(x-4)(x+5)$

Step-by-step explanation:

Each expression can be written as a product of linear factors as follows

1. X^2+1/2x ⇒ $x^{2} + \frac{1}{2}x$

$x^{2} + \frac{1}{2}x$ = $x(x+\frac{1}{2})$

Hence, X^2+1/2x = $x(x+\frac{1}{2})$

2. X^2+2x-3 ⇒ $x^{2} +2x-3$

$x^{2} +2x-3 = x^{2} +3x-x-3$

$x^{2} +3x-x-3 = x(x+3) -1(x+3)$

$x(x+3) -1(x+3) = (x-1)(x+3)$

Hence, X^2+2x-3 = $(x-1)(x+3)$

3. (2x-3)^2 ⇒ $(2x-3)^{2}$

$(2x-3)^{2} = (2x-3)(2x-3)$

Hence, (2x-3)^2 = (2x-3)(2x-3)

4. X^3+2x^2-19x-20 ⇒ $x^{3}+2x^{2} -19x -20$

$x^{3}+2x^{2} -19x -20 = (x+1)(x^{2} +x-20)$

First,

$x^{2} +x-20 = x^{2} +5x-4x-20\\x^{2} +5x-4x-20= x(x+5)-4(x+5)\\x(x+5)-4(x+5) = (x-4)(x+5)$

∴ $(x+1)(x^{2} +x-20) = (x+1)(x-4)(x+5)$

Hence, X^3+2x^2-19x-20 = $(x+1)(x-4)(x+5)$

3 0

3 years ago