Question

8 класс Чему равно "а", если остаток от деления х^4 - ах^3 + 4х^2 - х+1 на х-2 равен 7

Answer 1

По теореме Безу Теорема Безу утверждает, что остаток от деления многочлена P(x) на x-a  равен  P(a)
то есть  P(2)=16-8a+16-2+1=7
a=3

Answer 2

Solution for Finding the Value of "a"

To find the value of "a" when the remainder of the division of \(x^4 - ax^3 + 4x^2 - x + 1\) by \(x - 2\) is equal to 7, we can use the remainder theorem. According to the remainder theorem, if a polynomial \(f(x)\) is divided by \(x - c\), the remainder is \(f(c)\).

Using this theorem, we can set up the equation:

\(f(2) = 7\)

Where \(f(x) = x^4 - ax^3 + 4x^2 - x + 1\).

Let's calculate \(f(2)\) to find the value of "a".

Calculation

\(f(2) = 2^4 - a(2)^3 + 4(2)^2 - 2 + 1\)

\(f(2) = 16 - 8a + 16 - 2 + 1\)

\(f(2) = 31 - 8a\)

Now, we know that \(f(2) = 7\), so we can set up the equation:

\(31 - 8a = 7\)

Solving for "a":

\(31 - 8a = 7\)

\(-8a = 7 - 31\)

\(-8a = -24\)

\(a = -24 / -8\)

The value of "a" is 3.

Conclusion

After applying the remainder theorem and solving the equation, we find that the value of "a" is 3 when the remainder of the division of \(x^4 - ax^3 + 4x^2 - x + 1\) by \(x - 2\) is equal to 7.