Polynomial Functions: math test solutions, Grade 11 – download the PDF

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Test solutions with the detailed point scale. Add up your points and spot what to review.

Suggested time: 45 minutes. Out of 20 points. Calculator allowed only when the problem says so.

Sommaire

\(x^3 \div x = x^2\); \(x^2(x + 5) = x^3 + 5x^2\); subtract: \(-x^2 - 3x\) (1 pt).

\(-x^2 \div x = -x\); \(-x(x + 5) = -x^2 - 5x\); subtract: \(2x + 12\) (1 pt).

\(2x \div x = 2\); \(2(x + 5) = 2x + 10\); subtract: \(2\) (1 pt).

\(x^3 + 4x^2 - 3x + 12 = (x + 5)(x^2 - x + 2) + 2\) (1 pt).

With \(c = 3\): bring down \(2\); \(-9 + 6 = -3\); \(4 - 9 = -5\); \(15 - 15 = 0\). Quotient \(2x^2 - 3x - 5\), remainder \(0\) (2 pts).
The remainder is \(0\), so \(p(3) = 0\) and \(x - 3\) is a factor by the Factor Theorem (1 pt).
\(2x^2 - 3x - 5 = (2x - 5)(x + 1)\), so \(p(x) = (x - 3)(2x - 5)(x + 1)\) with zeros \(3\), \(\dfrac{5}{2}\) and \(-1\) (1 pt).

The degree is \(5\) (odd) and the leading coefficient is \(-2\) (negative), so the graph rises on the left and falls on the right (2 pts).
By the Fundamental Theorem of Algebra there are exactly \(5\) complex zeros, and there are at most \(5 - 1 = 4\) turning points (1 pt).

Divisors of \(3\): \(\pm 1, \pm 3\); divisors of \(2\): \(\pm 1, \pm 2\). Candidates: \(\pm 1, \pm 3, \pm\dfrac{1}{2}, \pm\dfrac{3}{2}\) (1 pt).
\(p(1) = 2 + 3 - 8 + 3 = 0\) (1 pt). Synthetic division gives \(2x^2 + 5x - 3 = (2x - 1)(x + 3)\), so the zeros are \(1\), \(\dfrac{1}{2}\) and \(-3\) (1 pt).