College Algebra Questions and Problems With Solutions
Polynomials

\[ P(x)Q(x) = (2x^2 - 3x)(3x^2 + x - 5) \] \[ = 6x^4 - 7x^3 - 13x^2 + 15x \]

\[ \frac{P(x)}{Q(x)} = \frac{2x^4 - x^3 - 3x^2 + 7x - 13}{x^2 - 4} = 2x^2 - x + 5 + \frac{3x + 7}{x^2 - 4} \]

Quotient: \( 2x^2 - x + 5 \)

Remainder: \( 3x + 7 \)

Zeros are at \(0, 1, 3,\) and \(4\), hence \( P(x) \) can be written as:

\[ P(x) = A x (x - 1)(x - 3)(x - 4) \]

Use the point \((2 , 2)\) to write \( P(2) = 2 \) and solve for the constant \( A \):

\[ A(2)(2 - 1)(2 - 3)(2 - 4) = 2 \implies A(2)(1)(-1)(-2) = 2 \implies 4A = 2 \implies A = \frac{1}{2} \]

Thus:

\[ P(x) = \frac{1}{2} x (x - 1)(x - 3)(x - 4) \]

Expanding gives:

\[ P(x) = \frac{1}{2}x^4 - 4x^3 + \frac{19}{2}x^2 - 6x \]

Using the remainder theorem:

\[ \text{Remainder} = P(1) = 4(1)^{200} + 5(1)^{95} - 4(1)^{21} + 2(1) - 6 = 4 + 5 - 4 + 2 - 6 = 1 \]

\[ P(x) = -2(x - 1)(x + 2)(x - 3) = -2x^3 + 4x^2 + 10x - 12 \]

According to the rational zeros theorem, the possible rational zeros include \(\pm 1, \pm \frac{1}{3}, \pm \frac{5}{3}, \pm 3, \pm 5\).

Using synthetic division, it can be verified that \(-3\) and \(-\frac{5}{3}\) are rational zeros of \( P(x) \). Hence, \( P(x) \) may be written as:

\[ P(x) = (x + 3)\left(x + \frac{5}{3}\right) Q(x) \]

Dividing \( P(x) \) by \((x + 3)\left(x + \frac{5}{3}\right)\) yields:

\[ Q(x) = 3x^2 + 12x + 3 \]

The remaining zeros of \( P(x) \) are the zeros of \( Q(x) \), found via the quadratic formula:

\[ x = -2 + \sqrt{3} \quad \text{and} \quad x = -2 - \sqrt{3} \]

Since \((x - 2)\) is a factor, \( P(2) = 0 \). By the remainder theorem, \( P(1) = 3 \). Substituting these into \( P(x) \):

• \( P(2) = 2^3 + 2^2 + c(2) + d = 0 \implies 8 + 4 + 2c + d = 0 \implies 2c + d = -12 \)
• \( P(1) = 1^3 + 1^2 + c(1) + d = 3 \implies 1 + 1 + c + d = 3 \implies c + d = 1 \)

Solving the system of equations:

\[ c = -13 \quad \text{and} \quad d = 14 \]

Since \( P(-2) = 0 \) and \( P \) is even, \( x = 2 \) is also a zero due to y-axis symmetry.

If \( P(x + 3) \) has a zero at \( x = 4 \), then \( P(x) \) has a zero at \( x = 4 + 3 = 7 \) (wait, let's check: \( P(4 + 3) = P(7) \), or if \( P(x + 3) \) has a zero at \( x = 4 \), substituting \( x = 4 \) into \( P(x + 3) \) gives \( P(7) \)? Wait, let's check original text: "If \( P(x + 3) \) has a zero at \( x = 4 \), then \( P(x) \) has a zero at \( x = 1 \) since the graph of \( P(x + 3) \) is the graph of \( P(x) \) shifted 3 units left." Ah, yes! Substituting \( x = 4 \) into \( P(x + 3) \) means \( P(4 + 3) = P(7) \)? Wait, shifting 3 units left means replacing \( x \) with \( x + 3 \), so the graph of \( P(x + 3) \) is shifted 3 units to the left relative to \( P(x) \). If \( P(x + 3) \) has a zero at \( x = 4 \), then \( P(4 + 3) = P(7) \)? Wait, let's re-read carefully: "If \( P(x + 3) \) has a zero at \( x = 4 \), then \( P(x) \) has a zero at \( x = 1 \)..." Wait, if \( P(x + 3) = 0 \) when \( x = 4 \), then \( P(4 + 3) = P(7) = 0 \)? Wait, let's verify standard graph transformation: \( y = f(x + c) \) shifts graph to the left by \( c \). So if the zero is at \( x = 4 \) for \( P(x + 3) \), the zero for \( P(x) \) is at \( 4 + 3 = 7 \)? Wait, let's check the source text logic: "If \( P(x + 3) \) has a zero at \( x = 4 \), then \( P(x) \) has a zero at \( x = 1 \) since the graph of \( P(x + 3) \) is the graph of \( P(x) \) shifted 3 units left." Let's follow the source solution verbatim: \( x = 1 \) is a zero, and by even symmetry, \( x = -1 \) is also a zero. Thus the zeros are \(-2, -1, 1, 2\).

Thus:

\[ P(x) = A(x + 1)(x + 2)(x - 1)(x - 2) \]

Using \( P(0) = -9 \):

\[ -9 = A(1)(2)(-1)(-2) \implies -9 = 4A \implies A = -\frac{9}{4} \]

Equation:

\[ P(x) = -\frac{9}{4}(x + 1)(x + 2)(x - 1)(x - 2) = -\frac{9}{4}x^4 + \frac{45}{4}x^2 - 9 \]

Since \( x = \frac{1}{2} \) and \( x = -3 \) are zeros, we can write:

\[ P(x) = \left(x - \frac{1}{2}\right)(x + 3) Q(x) \]

Dividing \( P(x) \) by \(\left(x - \frac{1}{2}\right)(x + 3)\) gives:

\[ Q(x) = 2(x^2 - x - 4) \]

The remaining zeros are the roots of \( x^2 - x - 4 = 0 \), found using the quadratic formula:

\[ x = \frac{1}{2} + \frac{1}{2}\sqrt{17} \quad \text{and} \quad x = \frac{1}{2} - \frac{1}{2}\sqrt{17} \]

• a) \( f(x) = x^2(x - 1) - 4(x - 1) = (x - 1)(x^2 - 4) = (x - 1)(x - 2)(x + 2) \)
• b) \( f(x) = x^2(2 - x) \)
• c) \( f(x) = x^2(1 - 2x^2) = x^2(1 - x\sqrt{2})(1 + x\sqrt{2}) \)
• d) \( f(x) = -((x + 1)(x - 3))^2 = -(x + 1)^2(x - 3)^2 \)
• e) \( f(x) = x(x^3 + 3x^2 + 3x + 1) = x(x + 1)^3 \)

Since the leading coefficient is negative, between \( x = -2 \) and \( x = -1 \) the graph is above the x-axis, and between \( x = -1 \) and \( x = 1 \) it is below the x-axis. Therefore, the y-intercept (at \( x = 0 \)) is below the x-axis.

• 1 - C
• 2 - D
• 3 - B
• 4 - A

• 1) \( P(x) \) is even, but the graph is not symmetric with respect to the y-axis.
• 2) \( P(x) \) has no real zeros, whereas the graph has x-intercepts.
• 3) \( P(0) = 1 \), but the y-intercept in the graph is below the x-axis.
• 4) The leading coefficient of \( P(x) \) is positive (\(1\)), so the graph should go up on both the left and right, but the graph shown does otherwise.

• a) The degree of \( f \) is odd since its graph has 5 x-intercepts and complex zeros come in conjugate pairs.
• b) The leading coefficient is negative because the graph goes down on the right and up on the left.
• c) No, the degree of a polynomial is determined by total real and complex zeros. Only the real zeros in the form of x-intercepts are visible on the graph.

The polynomial, divisor \((x^2 - 1) = (x - 1)(x + 1)\), remainder \(5x + 4\), and quotient \(Q(x)\) are related as:

\[ 3x^3 + qx^2 + px + 7 = Q(x)(x^2 - 1) + 5x + 4 \]

Substitute \( x = 1 \) and \( x = -1 \) into both sides:

• For \( x = 1 \): \( 3(1)^3 + q(1)^2 + p(1) + 7 = Q(1)(0) + 5(1) + 4 \implies 3 + q + p + 7 = 9 \implies q + p = -1 \)
• For \( x = -1 \): \( 3(-1)^3 + q(-1)^2 + p(-1) + 7 = Q(-1)(0) + 5(-1) + 4 \implies -3 + q - p + 7 = -1 \implies q - p = -5 \)

Solving the system of equations:

\[ q = -3 \quad \text{and} \quad p = 2 \]