Question

the movement of the progress bar may be uneven because questions can be worth more or less (including zero) depending on your answer. three polynomials are factored below, but some coefficients and constants are missing. all of the missing values of a, b, c, and d are integers. 1. $x^2 - 2x - 15 = (ax + b)(cx + d)$ 2. $2x^3 - 6x^2 - 36x = 2x(ax + b)(cx + d)$ 3. $4x^2 + 10x + 4 = (ax + b)(cx + d)$ fill in the table with the missing values of a, b, c, and d. | | a | b | c | d | |---|---|---|---|---| | 1. | 1 | -5 | 1 | | | 2. | 1 | | | -6 | | 3. | | 1 | 2 | |

Explanation:

Problem 1:

Step1: Expand the right side

$(ax + b)(cx + d)=acx^2+(ad + bc)x+bd$. For $x^2 - 2x - 15$, we know $a = 1$, $c = 1$, $b=-5$. So $ac = 1\times1 = 1$ (matches $x^2$ coefficient). Then $bd=1\times(-5)= - 5$? Wait, no, $bd$ should be $-15$ (constant term). Wait, $a = 1$, $c = 1$, $b=-5$, so $(x - 5)(x + d)=x^2+(d - 5)x-5d$. This should equal $x^2 - 2x - 15$. So equate coefficients:

Step2: Equate linear term and constant term

For linear term: $d - 5=-2\Rightarrow d = 3$. For constant term: $-5d=-15\Rightarrow d = 3$. So $d = 3$.

Problem 2:

Step1: Divide both sides by $2x$

We have $2x^3 - 6x^2 - 36x=2x(ax + b)(cx + d)$. Divide both sides by $2x$: $x^2 - 3x - 18=(ax + b)(cx + d)$. Given $a = 1$, $d=-6$, so $(x + b)(x - 6)=x^2+(b - 6)x-6b$. This should equal $x^2 - 3x - 18$.

Step2: Equate linear term and constant term

For linear term: $b - 6=-3\Rightarrow b = 3$. For constant term: $-6b=-18\Rightarrow b = 3$. Also, since $ac = 1$ (coefficient of $x^2$) and $a = 1$, then $c = 1$.

Problem 3:

Step1: Expand the right side

$(ax + b)(cx + d)=acx^2+(ad + bc)x+bd$. For $4x^2 + 10x + 4$, $b = 1$, $c = 2$. So $(ax + 1)(2x + d)=2ax^2+(ad + 2)x + d$.

Step2: Equate coefficients

For $x^2$ coefficient: $2a = 4\Rightarrow a = 2$. For constant term: $d\times1 = 4\Rightarrow d = 4$? Wait, no, $bd = 4$ (constant term) and $b = 1$, so $d = 4$? Wait, check linear term: $ad+bc=2d + 2\times1=2d + 2$. This should equal $10$. So $2d+2 = 10\Rightarrow 2d = 8\Rightarrow d = 4$. Let's verify: $(2x + 1)(2x + 4)=4x^2+8x + 2x + 4=4x^2 + 10x + 4$. Correct. So $a = 2$, $d = 4$.

Answer:

s:

1. $d = 3$
2. $b = 3$, $c = 1$
3. $a = 2$, $d = 4$

Filling the table:

	a	b	c	d
2	1	$\boldsymbol{3}$	$\boldsymbol{1}$	-6
3	$\boldsymbol{2}$	1	2	$\boldsymbol{4}$