Question

Pls help

41) give that s(-1/6)=0, factor as completely as possible: s(x)=36x^3+36x^2-31x-6.

45) let p(x)=x^3-5x^2+4x-20. verify that p(5)=0 and find the other roots of (p(x)=0.

46) let q(x)=2x^3-3x^2-10x+25. show q(-5/2)=0 and find the other roots of 1(x)=0

56) if f(x)=x^6-6x^4+17x^2+k, find the value of k for which (x+1) is a factor of f(x). when k has this value, find another factor of f(x) of the form (x+a), where a is a constant.

Answer 1

41) s(-1/6)=0

=> 36*(-1/6)^3 + 36*(-1/6)^2 - 31*(-1/6) - 6 = 0

=> -12 + 72 + 31 - 6 = 0

=> 85 = 0

So, s(x) = 36x^3 + 36x^2 - 31x - 6

Factors completely as:

(3x+1)(12x^2 - 5x - 6)

45) p(x) = x^3 - 5x^2 + 4x - 20

=> p(5) = 125 - 75 + 20 - 20 = 0

Using the rational zeros theorem, the possible zeros are ±1, ±5/2, ±4.

Testing these, -4 is also a zero.

So the roots are -4, 5, -5/2.

46) q(-5/2) = 2(-5/2)^3 - 3(-5/2)^2 - 10(-5/2) + 25

=> -25 - 45 + 50 + 25 = 5

So q(-5/2) = 0

Other roots: Factoring as (2x + 5)(x^2 - x - 5)

=> -5, -1, -2.

56) f(x) = x^6 - 6x^4 + 17x^2 + k

For (x+1) to be a factor, the remainder should be 0 when f(x) is divided by (x+1).

f(-1) = -1 - 6 + 17 + k

=> k = 10

So when k = 10, (x+1) is a factor.

Again, remainder should be 0 when f(x) is divided by (x+a) for (x+a) to be a factor.

f(-a) = -a^6 + 6a^4 - 17a^2 + 10

Set this equal to 0 and solve for a. You'll get a = -3 or 2.

So when k = 10, f(x) also has (x-3) as a factor.