practice multiple choice (highlight the correct answer)\n1. which of the following will cause the quantity…

Explanation:

Step1: Perform synthetic division for problem 9

Divide $4x^4 - 17x^3 - 27x^2 + 51x + 45$ by $(x - 5)$ using the root $x = 5$. $$5 \mid 4 \quad -17 \quad -27 \quad 51 \quad 45$$ $$\downarrow \quad 20 \quad 15 \quad -60 \quad -45$$ $$\overline{4 \quad 3 \quad -12 \quad -9 \quad 0}$$ The quotient is $4x^3 + 3x^2 - 12x - 9$.

Step2: Factor the quotient by grouping

Group the terms: $(4x^3 + 3x^2) - (12x + 9)$. $$x^2(4x + 3) - 3(4x + 3) = (x^2 - 3)(4x + 3)$$

Step3: Solve for roots of problem 9

Set each factor to zero: $x - 5 = 0$, $4x + 3 = 0$, and $x^2 - 3 = 0$. $$x = 5, \quad x = -\frac{3}{4}, \quad x = \pm\sqrt{3}$$

Step4: Perform synthetic division for problem 10

Divide $3x^4 + 15x^3 + 27x^2 + 56x + 60$ by $(x + 3)$ using the root $x = -3$. $$-3 \mid 3 \quad 15 \quad 27 \quad 56 \quad 60$$ $$\downarrow \quad -9 \quad -18 \quad -27 \quad -87$$ $$\overline{3 \quad 6 \quad 9 \quad 29 \quad -27}$$ Note: The image shows $x+3$ as a factor, but the remainder is not zero. Re-evaluating the image, the polynomial is $3x^4 + 15x^3 + 27x^2 + 12x + 20 = 0$ (based on the grouping shown). $$3x^3(x+5) + 4(3x+5) \text{ is incorrect. Let's use the grouping in the image: } (3x^4+15x^3) + (27x^2+12x+20)$$ Actually, the image shows: $x+3$ and $(3x^3+5x^2) + (12x+20)$. This implies the original was $3x^4 + 14x^3 + 27x^2 + 56x + 60$. However, following the grouping written: $$x^2(3x+5) + 4(3x+5) = (x^2+4)(3x+5)$$

Step5: Solve for roots of problem 10

Using the factors $(x+3)$, $(3x+5)$, and $(x^2+4)$. $$x = -3, \quad x = -\frac{5}{3}, \quad x = \pm 2i$$

Answer:

9. $x = 5, -\frac{3}{4}, \sqrt{3}, -\sqrt{3}$
10. $x = -3, -\frac{5}{3}, 2i, -2i$