Question

A tender brought $ 2{x^2} - x + 2 $ T.V. sets for Rs. $ 8{x^4} + 7x - 6 $ . Find the price of the T.V.

Answer 1

Hint: In this question it is give the cost of $ 2{x^2} - x + 2 $ T.V sets is Rs. $ 8{x^4} + 7x - 6 $ , so to find the value or price of one T.V set we have applied a unitary method.In this case we just divide the two algebraic expression to get the cost of T.V

Complete step-by-step answer:
Given, tender brought $ 2{x^2} - x + 2 $ T.V. sets for Rs. $ 8{x^4} + 7x - 6 $ .
So, the cost $ 2{x^2} - x + 2 $ televisions is Rs $ 8{x^4} + 7x - 6 $ .
Now use the unitary method as the cost $ 2{x^2} - x + 2 $ televisions is Rs $ 8{x^4} + 7x - 6 $ .
So, the cost of one television will be Rs $ \dfrac{{8{x^4} + 7x - 6}}{{2{x^2} - x + 2}} $ .

Now, we need to divide $ 8{x^4} + 7x - 6 $ by $ 2{x^2} - x + 2 $ .
  $
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{x^2} + 2x - 3 \\
  2{x^2} - x + 2\left| \!{\overline {\,
  \underline
  8{x^4} + 7x - 6 \\
  8{x^4} - 4{x^3} + 8{x^2} \\
    \\
  \,\,\,\,\,\,\,4{x^3} - 8{x^2} + 7x - 6 \\
  \,\,\,\,\,\,\,\underline {4{x^3} - 2{x^2} + 4x} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - 6{x^2} + 3x - 6 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline { - 6{x^2} + 3x - 6} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
  \,}} \right. \\
  $
Therefore, the cost of one T.V. will be Rs $ 4{x^2} + 2x - 3 $ .

Note: This question basically uses the concept of unitary method. In the unitary method we first find the unit value of any quantity, then we multiply the given quantity with unit quantity value.