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Pre-Algebra
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2a(a-5)+4(a-5)
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Hint: (a-5) is the common factor :)
Always take out the common factor :)
\[\huge 2a(a-5) + 4(a-5)\] Distribute. \[\huge 2a^2 - 10a + 4a - 20\] Simplify. \[\huge 2a^2 - 6a - 20\] \[\huge 2(a^2 -3a + 10)\] \[\huge 2(a-5)(a+2)\]
oops i changed the sign there
\[\huge 2(a^2 -3a - 10)\] \[\huge 2(a-5)(a+2)\] is the correct factor though
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lolz he's gone
yep
Why did you distribute @shamil98 ? kc_kennylau's method is only two steps 2a(a-5)+4(a-5) (2a+4)(a-5) 2(a+2)(a-5)
2a(a-5)+4(a-5) Multiply 2a by each term inside the parentheses. 2a^(2)-10a+4(a-5) Multiply 4 by each term inside the parentheses. 2a^(2)-10a+4a-20 Since -10a and 4a are like terms, subtract 4a from -10a to get -6a. 2a^(2)-6a-20
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