-2bk2+6bk-2b Final an outcome : -2b • (k2 - 3k + 1) action by action solution : step 1 :Equation in ~ the end of step 1 : ((0 - 2bk2) + 6bk) - 2b step 2 : action 3 :Pulling out favor terms : 3.1 ...

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-2r4+21r3-54r2 Final an outcome : -r2 • (r - 6) • (2r - 9) step by step solution : step 1 :Equation in ~ the finish of action 1 : ((0-(2•(r4)))+(21•(r3)))-(2•33r2) step 2 :Equation at the end of action 2 ...
-2kb2+6bk-2b Final an outcome : -2b • (kb - 3k + 1) action by action solution : action 1 :Equation in ~ the end of action 1 : ((0 - 2kb2) + 6kb) - 2b step 2 : step 3 :Pulling out like terms : 3.1 ...
George C. Might 17, 2015 notification that\displaystyle7=2+5and\displaystyle10=2\times5 .So we can factor: \displaystylea^2b^2-7ab+10=\left(ab\right)^2-7\left(ab\right)+10 ...
Alan P. Apr 9, 2015 Factoring by grouping is no a technique that is reasonable for an expression v 3 state (perhaps a fourth term to be intended(?))This expression,\displaystyle12a^3+20a^2-48a ...

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9k2+66k+21=0 Two solutions were uncovered : k = -7 k = -1/3 = -0.333 action by action solution : action 1 :Equation in ~ the end of action 1 : (32k2 + 66k) + 21 = 0 step 2 : step 3 :Pulling out ...
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\left< \beginarray l l 2 & 3 \\ 5 & 4 \endarray \right> \left< \beginarray together l together 2 & 0 & 3 \\ -1 & 1 & 5 \endarray \right>

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