Shresthakritagya59idn
Answered

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It is given that 2x³-7x²+x+10 = (x - 1) Q(j) + R. Find the remainder R and polynomial Q(j).​

Sagot :

Mhanifaidn

Answer:

  • 2x²- 5x - 4, remainder 6

Step-by-step explanation:

Divide the given by x - 1:

x - 1 | 2x³ - 7x² + x + 10 = 2x²- 5x - 4  rem 6

       | 2x³ - 2x²

               -5x² + x

              -5x² + 5x

                        - 4x + 10

                        - 4x + 4

                                  6

2x³-7x²+x+10 = (x - 1)(2x²- 5x - 4) + 6

Semsee45idn

Answer:

[tex]\large \text{$ Q(j) = 2x^2 - 5x - 4 $}[/tex]

[tex]\large \text{R = 6}[/tex]

Step-by-step explanation:

Use long division to find the polynomial and remainder:

[tex]\large \begin{array}{r}2x^2-5x-4\phantom{)}\\x-1{\overline{\smash{\big)}\,2x^3-7x^2+x+10\phantom{)}}}\\\underline{-~\phantom{(}(2x^3-2x^2)\phantom{-b)))))))}}\\0-5x^2+x+10\phantom{)}\\\underline{-~\phantom{()}(-5x^2+5x)\phantom{-b))}}\\0-4x+10\phantom{)}\\\underline{-~\phantom{()}(-4x+\phantom{)}4)}\\6\phantom{)}\end{array}[/tex]

Therefore:

[tex]\large \text{$2x^3 - 7x^2 + x + 10 = (x - 1)(2x^2 - 5x - 4) + 6$}[/tex]

So the polynomial and remainder are:

  • [tex]\large \text{$ Q(j) = 2x^2 - 5x - 4 $}[/tex]
  • [tex]\large \text{R = 6}[/tex]
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