Седми разред основне школе

Множење полинома 3

Решени задаци.

Задаци

Текст задатака објашњених у видео лекцији:

Пр.1)   Применом множења полинома испитати истинитост седећих

           формула:

           а) $\left( {x - 2} \right) \cdot \left( {{x^2} + 2x + 4} \right) = {x^3} - 8$

           б) $\left( {x + 1} \right) \cdot \left( {{x^3} - {x^2} + x - 1} \right) = {x^4} + 1$

Пр.2)   Решити једначине:

           а) $5x \cdot \left( {12x + 7} \right) - 4x \cdot \left( {15x - 11} \right) - 29x = 30$

           б) $\left( {x - 5} \right) \cdot \left( {x - 2} \right) - \left( {x + 4} \right) \cdot \left( {x - 1} \right) = 14$

           в) $ - 5\left( {3{x^2} - 2} \right) - \left( {5x - 1} \right) \cdot \left( { - 3x + 2} \right) = 14$

Пр.3)   Упростити израз, а затим израчунати његову бројевну вредност:

           а) $\left( {x - 4} \right) \cdot \left( {x - 2} \right) - \left( {x - 3} \right) \cdot \left( {x - 1} \right)$, ако је $x = 1$

           б) $\left( {\frac{1}{2}{x^2} + \frac{1}{4}x + \frac{1}{8}} \right) \cdot \left( {8x + 16{x^2}} \right)$, ако је $x =  - 1$.

Пр.4)   Ако је $P = 2{x^3} + 3{x^2} + 4x - 5$ и $Q = 3{x^3} + 2{x^2} - 3x - 1$,

           израчунати $3P - 2Q$, а затим израчунати вредност добијеног

           израза за $x =  - 2$.

 

Пр.1)   

а)

\[\begin{gathered}
\left( {x - 2} \right) \cdot \left( {{x^2} + 2x + 4} \right) = {x^3} - 8 \hfill \\
\left( {x - 2} \right) \cdot \left( {{x^2} + 2x + 4} \right) = {x^3} + 2{x^2} + 4x - 2{x^2} - 4x - 8 = {x^3} - 8 \hfill \\
{x^3} - 8 = {x^3} - 8 \hfill \\
\end{gathered} \]

б) 

\[\begin{gathered}
\left( {x + 1} \right) \cdot \left( {{x^3} - {x^2} + x - 1} \right) = {x^4} + 1 \hfill \\
\left( {x + 1} \right) \cdot \left( {{x^3} - {x^2} + x - 1} \right) = {x^4} - {x^3} + {x^2} - x + {x^3} - {x^2} + x - 1 = {x^4} - 1 \hfill \\
{x^4} - 1 \ne {x^4} + 1 \hfill \\
\end{gathered} \]

 

Пр.2)   

а)

\[\begin{gathered}
5x \cdot \left( {12x + 7} \right) - 4x \cdot \left( {15x - 11} \right) - 29x = 30 \hfill \\
60{x^2} + 35x - 60{x^2} + 44x - 29x = 30 \hfill \\
50x = 30 \hfill \\
x = \frac{3}{5} \hfill \\
\end{gathered} \]

б) 

\[\begin{gathered}
\left( {x - 5} \right) \cdot \left( {x - 2} \right) - \left( {x + 4} \right) \cdot \left( {x - 1} \right) = 14 \hfill \\
{x^2} - 2x - 5x + 10 - \left( {{x^2} - x + 4x - 4} \right) = 14 \hfill \\
{x^2} - 2x - 5x + 10 - {x^2} + x - 4x + 4 = 14 \hfill \\
- 10x + 14 = 14 \hfill \\
- 10x = 0 \hfill \\
x = 0 \hfill \\
\end{gathered} \]

в) 

\[\begin{gathered}
- 5\left( {3{x^2} - 2} \right) - \left( {5x - 1} \right) \cdot \left( { - 3x + 2} \right) = 14 \hfill \\
- 15{x^2} + 10 - \left( { - 15{x^2} + 10x + 3x - 2} \right) = 14 \hfill \\
- 15{x^2} + 10 + 15{x^2} - 10x - 3x + 2 = 14 \hfill \\
- 13x + 12 = 14 \hfill \\
- 13x = 2 \hfill \\
x = - \frac{2}{{13}} \hfill \\
\end{gathered} \]

 

Пр.3)   

а) 

\[\begin{gathered}
\left( {x - 4} \right) \cdot \left( {x - 2} \right) - \left( {x - 3} \right) \cdot \left( {x - 1} \right) = {x^2} - 2x - 4x + 8 - {x^2} + x + 3x - 3 = \hfill \\
= - 2x + 5 \hfill \\
- 2 \cdot 1 + 5 = - 2 + 5 = 3 \hfill \\
\end{gathered} \]

б) 

\[\begin{gathered}
\left( {\frac{1}{2}{x^2} + \frac{1}{4}x + \frac{1}{8}} \right) \cdot \left( {8x + 16{x^2}} \right) = 4{x^3} + 8{x^4} + 2{x^2} + 4{x^3} + x + 2{x^2} = \hfill \\
= 8{x^4} + 8{x^3} + 4{x^2} + x \hfill \\
8{\left( { - 1} \right)^4} + 8{\left( { - 1} \right)^3} + 4{\left( { - 1} \right)^2} + \left( { - 1} \right) = 8 - 8 + 4 - 1 = 3 \hfill \\
\end{gathered} \]

 

Пр.4)   

\[\begin{gathered}
3 \cdot \left( {2{x^3} - 3{x^2} + 4x - 5} \right) - 2\left( {3{x^3} + 2{x^2} - 3x - 1} \right) = \hfill \\
= 6{x^3} - 9{x^2} + 12x - 15 - 6{x^3} - 4{x^2} + 6x + 2 = - 13{x^2} + 18x - 13 \hfill \\
\hfill \\
- 13{\left( { - 2} \right)^2} + 18\left( { - 2} \right) - 13 = - 13 \cdot 4 - 36 - 13 = - 52 - 36 - 13 = - 101 \hfill \\
\end{gathered} \]