Question

1. a) what value of b will ensure x + 3 is a factor of ( bx^3 - 2x^2 + x - 6 )?

b) what value of d will ensure x + 2 is a factor of ( 3x^5 - dx^4 + 4x^3 - 2dx^2 + x + 10 )?

Explanation:

Step1: Solve part (a): Find root

If $x+3$ is a factor, then $x=-3$ is a root. Substitute $x=-3$ into $bx^3 - 2x^2 + x - 6 = 0$.
$$b(-3)^3 - 2(-3)^2 + (-3) - 6 = 0$$

Step2: Simplify the equation

Calculate each term and combine like terms.
$$-27b - 18 - 3 - 6 = 0$$
$$-27b - 27 = 0$$

Step3: Solve for b

Isolate the variable $b$.
$$-27b = 27$$
$$b = \frac{27}{-27} = -1$$

Step4: Solve part (b): Find root

If $x+2$ is a factor, then $x=-2$ is a root. Substitute $x=-2$ into $3x^5 - dx^4 + 4x^3 - 2dx^2 + x + 10 = 0$.
$$3(-2)^5 - d(-2)^4 + 4(-2)^3 - 2d(-2)^2 + (-2) + 10 = 0$$

Step5: Simplify the equation

Calculate each term and combine like terms.
$$3(-32) - d(16) + 4(-8) - 2d(4) + 8 = 0$$
$$-96 - 16d - 32 - 8d + 8 = 0$$
$$-120 - 24d = 0$$

Step6: Solve for d

Isolate the variable $d$.
$$-24d = 120$$
$$d = \frac{120}{-24} = -5$$

Answer:

a) $b = -1$
b) $d = -5$