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Troyanec [42]
2 years ago
12

In factored form please help

Mathematics
1 answer:
Brilliant_brown [7]2 years ago
3 0

Answer:

3rd option

Step-by-step explanation:

\frac{3x^2-3}{x^2-5x+4} ( factorise numerator and denominator )

3x² - 3 ← factor out 3 from each term

= 3(x² - 1²) ← x² - 1 is a difference of squares and factors in general as

a² - b² = (a - b)(a + b)

x² - 1

= x² - 1²

= (x - 1)(x + 1) , then

3x² - 3 = 3²(x - 1)(x + 1) ← in factored form

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

x² - 5x + 4

consider the factors of the constant term (+ 4) which sum to give the coefficient of the x- term (- 5)

the factors are - 1 and - 4 , since

- 1 × - 4 = + 4 and - 1 - 4 = - 5 , then

x² - 5x + 4 = (x - 1)(x - 4)

then

\frac{3x^2-3}{x^2-5x+4} = \frac{3(x-1)(x+1)}{(x-1)(x-4)} ← in factored form

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Select the correct answer. Which expression is equivalent to 8x^2^3 sqrt 375x + 2^3 sqrt 3x^7, if x=0?
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Answer:

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Step-by-step explanation:

The question is poorly formatted. The original question is:

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7

We have:

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7

Open bracket

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^{3* \frac{2}{3}}) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(8^ \frac{2}{3} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

Express 8 as 2^3

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^{3* \frac{2}{3}} *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =(2^2 *x^2) \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7} =4x^2 \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}

Express 2^3 as 8

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{75x^3} + 8\sqrt{ 3x^7}

Expand each exponent

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2 *3x} + 8\sqrt{ 3x * x^6}

Split

(8x^3)^ \frac{2}{3} \sqrt{75x^3} + 2^3 \sqrt{ 3x^7}=4x^2 \sqrt{25x^2} *\sqrt{3x} + 8\sqrt{3x} * \sqrt{x^6}

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Factorize

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