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2 years ago

Solve using perfect square factoring patterns. y2 + 14y + 49 = 0

1 answer:

4 0

Y² + 14y + 49 = 0

We know a² + 2ab + b² = (a + b)²
(y)² + 2 x 7 x y + (7)² = 0
(y + 7)² = 0
y = -7

y = -7

Correct option C

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2 years ago

Tamara and Clyde got different answers when dividing 2x4 + 7x3 – 18x2 + 11x – 2 by 2x2 – 3x + 1. Analyze their individual work.

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<em>Note: As you may have unintentionally missed to add the different answers, based on which we had to check who solved correctly between Tamara and Clyda's work. </em>

<em>But, I am actually solving the expression and you must note that whoever (between Tamara and Clyda's work) may have got the same answer or match the answer with mine, would be the one who solved correctly.</em>

Answer:

We conclude that whoever (between Tamara and Clyda's work) may have got the answer as $x^2+5x-2$ after dividing $2x^4\:+\:7x^3\:-\:18x^2\:+\:11x\:-\:2$ by $2x^2\:-\:3x\:+\:1$ , would be the one who solved it correctly.

Step-by-step explanation:

Considering the expression

$2x^4\:+\:7x^3\:-\:18x^2\:+\:11x\:-\:2$

Lets divide the expression by $2x^2\:-\:3x\:+\:1$

Solution Steps:

$\frac{2x^4+7x^3-18x^2+11x-2}{2x^2-3x+1}$

Factorizing

$2x^4+7x^3-18x^2+11x-2:\quad \left(x-1\right)\left(2x-1\right)\left(x^2+5x-2\right)$

$\frac{\left(x-1\right)\left(2x-1\right)\left(x^2+5x-2\right)}{2x^2-3x+1}$

Factorizing

$2x^2-3x+1:\quad \left(2x-1\right)\left(x-1\right)$

$\frac{\left(x-1\right)\left(2x-1\right)\left(x^2+5x-2\right)}{\left(2x-1\right)\left(x-1\right)}$

$\mathrm{Cancel\:}\frac{\left(x-1\right)\left(2x-1\right)\left(x^2+5x-2\right)}{\left(2x-1\right)\left(x-1\right)}:\quad x^2+5x-2$

$x^2+5x-2$

Thus,

$\frac{2x^4+7x^3-18x^2+11x-2}{2x^2-3x+1}=x^2+5x-2$

Therefore, we conclude that whoever (between Tamara and Clyda's work) may have got the answer as $x^2+5x-2$ after dividing $2x^4\:+\:7x^3\:-\:18x^2\:+\:11x\:-\:2$ by $2x^2\:-\:3x\:+\:1$ , would be the one who solved it correctly.

Keywords: expression, division

Learn more about expression division from brainly.com/question/1575482

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Step-by-step explanation:

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