Algebra · real student question

Simplify (x + 1) x 0.79 x 1.15 x 1.3 x 1.3 x 1.15.

Question

Simplify (x+1)⋅0.79⋅1.15⋅1.3⋅1.3⋅1.15(x+1)\cdot 0.79 \cdot 1.15 \cdot 1.3 \cdot 1.3 \cdot 1.15.

Step-by-step solution

  1. Separate the variable part from the constants. Multiplication is commutative and associative, so the (x+1)(x+1) can be set aside and all five decimals collapsed into one number: (x+1)⋅(0.79⋅1.15⋅1.3⋅1.3⋅1.15)(x+1)\cdot\big(0.79\cdot1.15\cdot1.3\cdot1.3\cdot1.15\big). This is the whole point of the exercise - a chain of five successive multipliers is equivalent to a single one.

  2. Group the repeated factors as squares. Two of each repeat, so 1.15⋅1.15=1.152=1.3225,1.3⋅1.3=1.32=1.69.1.15\cdot1.15 = 1.15^2 = 1.3225, \qquad 1.3\cdot1.3 = 1.3^2 = 1.69. Squaring is quicker and less error-prone than four separate multiplications.

  3. Multiply the three remaining constants. 0.79⋅1.3225=1.044775,1.044775⋅1.69=1.76566975.0.79 \cdot 1.3225 = 1.044775, \qquad 1.044775 \cdot 1.69 = 1.76566975. Work left to right and keep every digit; each factor has at most four decimals, so the exact product has at most eight.

  4. Confirm the product is exact, not rounded. As fractions, 0.79⋅529400⋅169100=70626794000000=1.765669750.79\cdot\tfrac{529}{400}\cdot\tfrac{169}{100} = \tfrac{7062679}{4000000} = 1.76566975 exactly, so no precision has been lost.

  5. Write the simplified expression. (x+1)⋅0.79⋅1.15⋅1.3⋅1.3⋅1.15=1.76566975 (x+1).(x+1)\cdot 0.79\cdot1.15\cdot1.3\cdot1.3\cdot1.15 = 1.76566975\,(x+1).

  6. Distribute only if a linear form is needed. 1.76566975(x+1)=1.76566975x+1.76566975.1.76566975(x+1) = 1.76566975x + 1.76566975. The factored form is usually more useful; the expanded form is what you want if the expression must be added to another polynomial.

Answer

1.76566975 (x+1)=1.76566975x+1.765669751.76566975\,(x+1) = 1.76566975x + 1.76566975

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