Algebra Polynomial Divisibility and Systems of Equations

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Questions and Answers

What must be the value of $m$ in the polynomial $P(x) = 4x^{3} - 2x^{2} + mx - 2$ for it to leave a remainder of 1 when divided by $(x-1)$?

  • 3 (correct)
  • 1
  • 7
  • 5

Which stated procedure can tell if a polynomial $P(x)$ is divisible by $(x-a)$ without division?

  • Determining the leading coefficient
  • Finding $P(a)$ and checking if it equals zero (correct)
  • Calculating the roots of the polynomial
  • Examining the degree of the polynomial

What is the solution for $x$ and $y$ in the system $egin{cases} x - y = 1 \ 2x - 2y = 4 ext{?} \

  • (1, 0)
  • (4, 3)
  • (2, 1) (correct)
  • (3, 2)

For the system of equations $egin{cases} logx - logy = 1 \ x + y = 22 \ $

<p>(10, 12) (D)</p> Signup and view all the answers

What solution satisfies the system $egin{cases} 3x + y = 1 \ xy = -2 \ $?

<p>(1, -3) (A)</p> Signup and view all the answers

Flashcards

Divisibility of Polynomials

A polynomial P(x) is divisible by (x-a) if substituting x=a into P(x) results in 0.

Remainder Theorem

When a polynomial P(x) is divided by (x-a), the remainder is P(a).

System of Equations

A set of two or more equations with the same variables that need to be solved simultaneously.

Consistent System (Example a)

Equations x - y = 1 and 2x - 2y = 4 have the same solution set.

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Inconsistent System (Example a)

If the equations have no common solution.

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Study Notes

Polynomial Divisibility

  • To determine if a polynomial P(x) is divisible by another polynomial (x-a) without division, evaluate P(a). If the result is zero, (x-a) is a factor.

  • For polynomial P(x) = 4x³ - 2x² + mx - 2, if division by (x-1) leaves a remainder of 1, then P(1) = 1.

Systems of Equations

  • Problem a: Solving the system x - y = 1, 2x - 2y = 4. This system has no unique solution.

  • Problem b: Solving the system logx - logy = 1, x + y = 22. This involves logarithmic rules and substitution.

  • Problem c: Solving the system 3x + y = 1, xy = -2. This system involves substitution and possible quadratic equations.

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