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3 years ago

Factor trinomial by grouping : 10x^2-9x+2

Mathematics

1 answer:

Angelina_Jolie [31]3 years ago

8 0

Answer:

(2x - 1)(5x - 2)

Step-by-step explanation:

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term, that is

product = 10 × 2 = 20 and sum = - 9

The factors are - 5 and - 4

Split the middle term using these factors

10x² - 5x - 4x + 2 (factor the first/second and third/fourth terms by grouping )

= 5x(2x - 1) - 2(2x - 1) ← factor out (2x - 1)

= (2x - 1)(5x - 2)

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Perform the multiplication. Simplify the answers. 2 √30*( √5+ √6+ √10+ √15)

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The simplified expression of $2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15})$ is $10\sqrt{6} + 6\sqrt{20}+ 20\sqrt{3} + 30\sqrt{2}$

The expression is given as:

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15})$

Expand the expression

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2\sqrt{30} \times \sqrt5+ 2\sqrt{30} \times \sqrt6+ 2\sqrt{30} \times \sqrt{10} + 2\sqrt{30} \times \sqrt{15}$

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$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(\sqrt{30} \times \sqrt5+ \sqrt{30} \times \sqrt6+ \sqrt{30} \times \sqrt{10} + \sqrt{30} \times \sqrt{15})$

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$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(\sqrt{150} + \sqrt{180}+ \sqrt{300} + \sqrt{450})$

Expand the expression

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Evaluate the roots

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) = 2(5\sqrt{6} + 3\sqrt{20}+ 10\sqrt{3} + 15 \sqrt{2})$

Expand

$2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15}) =10\sqrt{6} + 6\sqrt{20}+ 20\sqrt{3} + 30\sqrt{2}$

Hence, the simplified expression of $2\sqrt{30} \times (\sqrt5+ \sqrt6+ \sqrt{10} + \sqrt{15})$ is $10\sqrt{6} + 6\sqrt{20}+ 20\sqrt{3} + 30\sqrt{2}$

Read more about simplified expressions at:

brainly.com/question/8008182

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Factor trinomial by grouping : 10x^2-9x+2​

Factor trinomial by grouping : 10x^2-9x+2