- Identify the value \(c\) from the divisor \(x-c\). - The remainder theorem replaces the division calculation with one function evaluation. - Substitute carefully into every term.

1. For divisor \(x-2\), the remainder theorem says the remainder is \(f(2)\). 2. Compute \(f(2)=2(2)^3-(2)^2+4(2)-5=15\).

The remainder is \(15\).

- Read the x-value where the right side of the graph meets the x-axis. - What linear expression becomes zero at that input?

1. The right-hand x-intercept is at \(x=2\), where \(f(2)=0\). 2. By the factor theorem, \(x-2\) is a factor.

\(x-2\)

- For a divisor of the form \(x-c\), which function value gives the remainder? - Which number is \(c\) here?

1. The remainder theorem says the remainder on division by \(x-1\) is \(P(1)\). 2. Evaluate \(P(1)=2-3+4=3\).

\(3\)

- What factor is linked to a zero \(r\)? - Which of the two choices actually equals zero at \(x=-4\)?

1. A zero at \(x=-4\) gives a factor of the form \(x-(-4)\). 2. Simplify this factor to \(x+4\).

\(x+4\), because it equals zero when \(x=-4\).

- Which input makes the divisor \(x+1\) equal zero? - Find the function value for that input in the table.

1. Rewrite the divisor as \(x-(-1)\). 2. The remainder is \(P(-1)\), which the table gives as \(-2\).

\(-2\)

- What does a remainder of zero say about divisibility? - How is exact divisibility related to factors?

1. A zero remainder means the division is exact. 2. Therefore, \(x-7\) is a factor of \(P(x)\).

\(x-7\) is a factor of \(P(x)\).

- A divisor \(x-c\) is a factor exactly when the polynomial evaluates to \(0\) at \(c\). - Pay attention to the sign in \(x+3\) when identifying the test value. - You do not need to calculate a quotient.

1. For \(x+3=x-(-3)\), evaluate \(p(-3)=-27+18+15-6=0\). By the factor theorem, \(x+3\) is a factor. 2. For \(x-1\), evaluate \(p(1)=1+2-5-6=-8\ne0\). Therefore, \(x-1\) is not a factor.

a) Yes, \(x+3\) is a factor. b) No, \(x-1\) is not a factor.

- Rewrite the divisor in the form \(x-c\) before choosing the input value. - The sign of \(c\) matters when the divisor is \(x+2\). - Evaluate the polynomial directly instead of dividing it.

1. Since \(x+2=x-(-2)\), the remainder is \(g(-2)\). 2. Compute \(g(-2)=(-2)^4-3(-2)^3+2(-2)-7=16+24-4-7=29\).

The remainder is \(29\).

- Rewrite each divisor in the form \(x-c\). - For \(x+2\), decide which signed value makes that divisor zero. - How does changing the input from positive to negative affect the odd-power terms?

1. For \(x-2\), evaluate \(f(2)=8-8-10+6=-4\). The remainder is \(-4\). 2. Since \(x+2=x-(-2)\), evaluate \(f(-2)=-8-8+10+6=0\). The remainder is \(0\), so \(x+2\) is a factor. 3. The remainder theorem uses \(c\) for a divisor written as \(x-c\): the test inputs are \(2\) and \(-2\), respectively.

a) \(-4\) b) \(0\) c) \(x-2=x-(2)\), but \(x+2=x-(-2)\), so the inputs are \(2\) and \(-2\).

- Translate “\(x-2\) is a factor” into a statement about \(p(2)\). - Substitute before solving for the unknown coefficient. - The factor theorem turns the factor condition into one linear equation in \(k\).

1. If \(x-2\) is a factor, then \(p(2)=0\). 2. Compute \(p(2)=8+4k-10+6=4k+4\). 3. Solve \(4k+4=0\) to get \(k=-1\).

\(k=-1\)

- Recall exactly which part of polynomial division the remainder theorem identifies. - A quotient is a polynomial, while the theorem gives one number. - Consider what a nonzero remainder says about factor status.

1. The remainder theorem states that \(p(4)\) is the remainder when dividing by \(x-4\), not the quotient. 2. Therefore, the remainder is \(57\). 3. Because the remainder is nonzero, \(x-4\) is not a factor of \(p(x)\). The theorem alone does not determine the quotient.

Mireille confused the quotient with the remainder. The remainder is \(57\), so \(x-4\) is not a factor; the quotient is not determined by \(p(4)\) alone.

- Ask whether the problem needs a quotient or only a factor decision. - Each linear factor candidate determines one input value to test. - A zero function value is the deciding condition.

1. The factor theorem is more efficient because only factor status is needed. 2. Evaluate \(f(1)=3-2+1-2=0\). Therefore, \(x-1\) is a factor. 3. Evaluate \(f(-1)=-3+2-1-2=-4\ne0\). Therefore, \(x+1\) is not a factor.

Use the factor theorem. \(x-1\) is a factor, and \(x+1\) is not.

- What input makes the divisor zero? - Use that input in every term of the polynomial. - Pay attention to the sign of an odd power of a negative number.

1. Since \(x+2=x-(-2)\), evaluate \(P(-2)\). 2. Calculate \(3(-2)^3+(-2)^2-2=-24+4-2=-22\).

\(-22\)

- Write the factor associated with a zero of \(-3\) before simplifying signs. - Which factor actually becomes zero when you substitute \(-3\)? - Test the other factor with its own zero to rule it out.

1. Substitute \(-3\): \(P(-3)=9-6-3=0\). 2. A zero at \(-3\) gives the factor \(x+3\), not \(x-3\). 3. In fact, \(P(3)=9+6-3=12\), so \(x-3\) is not a factor.

Jordan used the wrong sign. \(P(-3)=0\) shows that \(x+3\) is a factor; \(x-3\) is not.

- What function value must equal zero when \(x+2\) is a factor? - Substitute that input while leaving \(a\) as an unknown. - Solve the equation for the unknown coefficient and check it.

1. Because \(x+2\) is a factor, \(f(-2)=0\). 2. Substitute to get \(-8-2a+4=0\), or \(-4-2a=0\). 3. Solve to find \(a=-2\).

Since \(x+2\) is a factor, \(f(-2)=0\). Thus \(-8-2a+4=0\), so \(a=-2\).

- For division by \(x-c\), the remainder theorem connects the remainder directly to \(P(c)\). - A linear expression \(x-c\) is a factor exactly when the corresponding function value is \(0\). - For \(z+2\), identify the value of \(c\) in the form \(z-c\). - After confirming the factor in part 2, divide once and factor the quadratic quotient.

1) By the remainder theorem, the remainder on division by \(x-1\) is \(P(1)\). Compute \(P(1)=1+3+4=8\). The remainder is \(8\), so \(x-1\) is not a factor. 2) Because \(z+2=z-(-2)\), evaluate \(Q(-2)=(-2)^3-6(-2)^2+32=-8-24+32=0\). By the factor theorem, \(z+2\) is a factor. Use synthetic value \(-2\) with coefficients \(1,-6,0,32\): bring down \(1\); multiply by \(-2\) and add to get \(-8\); multiply by \(-2\) and add to get \(16\); multiply by \(-2\) and add to get remainder \(0\). The quotient is \(z^2-8z+16\). Since \(z^2-8z+16=(z-4)^2\), the complete factorization is \(Q(z)=(z+2)(z-4)^2\).

1) Remainder \(8\); \(x-1\) is not a factor. 2) \(Q(-2)=0\), so \(z+2\) is a factor, and \(Q(z)=(z+2)(z-4)^2\).

- Use the given zero in the polynomial to determine \(q\). - After finding \(q\), factor the resulting quadratic. - Set each linear factor equal to zero to identify the zeros.

1. Since \(x = -1\) is a zero, \(P(-1) = 0\). 2. Substitute \(x = -1\): \((-1)^2 + 5(-1) + q = 0\). 3. Simplify: \(1 - 5 + q = 0\), so \(q = 4\). 4. Then \(P(x) = x^2 + 5x + 4 = (x + 1)(x + 4)\). 5. Therefore, the zeros are \(x = -1\) and \(x = -4\).

\(q = 4\); \(P(x) = (x + 1)(x + 4)\); zeros: \(x = -1\) and \(x = -4\)

- Use the factor theorem to connect the factor \(x + 2\) with the zero \(-2\). - Substitute the zero to determine \(b\). - Divide or compare coefficients to find the remaining factor.

1. If \(x + 2\) is a factor, then \(x = -2\) is a zero. 2. Substitute \(x = -2\): \(3(-2)^2 + b(-2) - 10 = 0\). 3. Simplify: \(12 - 2b - 10 = 0\), so \(b = 1\). 4. The expression is \(3x^2 + x - 10\). Since one factor is \(x + 2\), the other factor is \(3x - 5\). 5. Therefore, \(3x^2 + x - 10 = (x + 2)(3x - 5)\).

1. \(b = 1\) 2. \((x + 2)(3x - 5)\)

- A prescribed remainder gives a prescribed function value. - Substitute the divisor's zero into the polynomial before solving for \(k\). - Do not set the function value to \(0\) unless the problem says the divisor is a factor.

1. The remainder theorem gives \(p(2)=7\). 2. Compute \(p(2)=16+4k-6+1=4k+11\). 3. Solve \(4k+11=7\), giving \(4k=-4\) and \(k=-1\).

\(k=-1\)

- Convert the factor condition and the remainder condition into two function-value equations. - Be careful that \(x+2\) corresponds to the input \(-2\). - Solve the resulting two linear equations for the two unknown coefficients.

1. Since \(x-1\) is a factor, \(p(1)=0\), so \(1+a+b=0\). 2. The divisor \(x+2\) corresponds to \(x-(-2)\), so \(p(-2)=9\). This gives \(-8-2a+b=9\). 3. From the first equation, \(a+b=-1\). From the second, \(-2a+b=17\). 4. Subtract the first equation from the second: \(-3a=18\), so \(a=-6\). Then \(b=5\).

\(a=-6\) and \(b=5\)

- The remainder theorem gives the remainder from one evaluation. - For the verification, use the same value in synthetic division. - Compare the final synthetic entry with the theorem result.

1. By the remainder theorem, the remainder is \(f(3)=27-18+12+1=22\). 2. Use synthetic value \(3\) with coefficients \(1,-2,4,1\). Bring down \(1\), then multiply by \(3\) and add to get \(1\), \(7\), and remainder \(22\). 3. The quotient is \(x^2+x+7\), and the synthetic remainder agrees with \(f(3)\).

a) \(22\) b) Synthetic division with \(3\) gives bottom row \(1,1,7,22\), ending in remainder \(22\).

- Start from the polynomial division identity given in the problem. - Substitute the value that makes the divisor \(x-c\) equal to \(0\). - Connect “is a factor” with having remainder \(0\).

1. Substitute \(x=c\) into the division identity: \(p(c)=(c-c)q(c)+r=0\cdot q(c)+r=r\). 2. Thus, the function value \(p(c)\) equals the remainder. 3. The divisor \(x-c\) is a factor exactly when the remainder is \(0\). Since \(p(c)=r\), this happens exactly when \(p(c)=0\).

Substituting \(x=c\) gives \(p(c)=r\). Therefore, \(x-c\) is a factor exactly when \(r=0\), which is equivalent to \(p(c)=0\).

- Translate the factor condition and the prescribed remainder into two evaluations. - Determine the unknown coefficients before trying to factor the polynomial. - Once the coefficients are known, use the given factor to reduce the cubic.

1. Since \(x-2\) is a factor, \(p(2)=0\): \(8+4a+2b+6=0\), so \(2a+b=-7\). 2. Since the remainder on division by \(x+1\) is \(12\), \(p(-1)=12\): \(-1+a-b+6=12\), so \(a-b=7\). 3. Solve the equations to obtain \(a=0\) and \(b=-7\). 4. Then \(p(x)=x^3-7x+6\). Dividing by the known factor \(x-2\) gives \(x^2+2x-3=(x-1)(x+3)\), so \(p(x)=(x-2)(x-1)(x+3)\).

a) \(a=0\), \(b=-7\) b) \(p(x)=(x-2)(x-1)(x+3)\)

- Find the height of the curve above or below \(x=1\). - Once you know the formula, what input tests divisibility by \(x-3\)? - The factor theorem lets you test a factor even when its zero lies outside the graph window.

1. At \(x=1\), the graph has value \(-10\). Since \(p(1)=1-1-4-6=-10\) but \(q(1)=1-1-2-6=-8\), the graph represents \(p\). 2. By the factor theorem, \(x-3\) is a factor if \(p(3)=0\). Here \(p(3)=27-9-12-6=0\), so it is a factor. The root at \(3\) is outside the displayed window.

a) \(p(x)=x^3-x^2-4x-6\), because the graph passes through \((1,-10)\) b) Yes. \(p(3)=0\), so \(x-3\) is a factor.

- What value of \(d(t)\) means the robot is exactly at the marker? - Connect this zero to a linear factor, then use the same input to find the unknown coefficient. - Once \(k\) is known, factor out the common variable before finding the other zeros.

1. At the marker, \(d(2)=0\), so the factor theorem gives the factor \(t-2\). 2. Substitute \(t=2\) into the polynomial: \(8-24+2k=0\), so \(k=8\). 3. Factor: \(d(t)=t(t^2-6t+8)=t(t-2)(t-4)\). 4. All zeros in the interval occur at \(t=0\), \(t=2\), and \(t=4\), measured in seconds.

a) \(t-2\), since \(d(2)=0\) b) \(k=8\) c) \(d(t)=t(t-2)(t-4)\); at the marker at \(0\), \(2\), and \(4\) seconds