Question

What polynomial has roots of -6, -4, and 1 ?

x3 - 9x2 - 22x + 24

x3 - x2 - 26x - 24

x3 + x2 - 26x + 24

x3 + 9x2 + 14x - 24

Answer 1

Have you ever done this type of problems before? If what, what did you do ?

Answer 2

(x+6)(x+4)(x-1) = 0

Answer 3

ok then what do i do? @cwrw238

Answer 4

Would I foil @cwrw238

Answer 5

you can foil (x+4)(x-1) if you like then multiply this by (x + 6)

Answer 6

this will give you
(x + 6)(x^2 + 3x - 4)
now use the distributive law
multiply the second bracket by x then by + 6 and simplify
to give one of the above options

Answer 7

well an easier method may be to use the sum and product of the roots of a cubic

\[ax^3 + bx^2 + cx + d = 0\]

then using the connection between roots and coefficients

sum of the roots

\[-\frac{b}{a} = -6 + (-4) + 1 \]

in this question a = 1 so

-b = -9

hence b = 9

then the sum of the porduct of pairs

\[\frac{c}{a} = -6 \times -4 + -6 \times 1 + -4 \times 1 = 14\]

then c = 14

and lastly the product of the roots

\[\frac{-d}{a} = -4 \times -6 \times 1\]

-d = 24

so d = -24

so the substitute a = 1, b = -9 c = 14 and d = -24 into the general form at the top

Answer 8

General method: substitute these values: -6, -4, 1 into these expressions to see which ones equal zero.
Only the third expression has (1) as root.
Only the second expression has (-4) as root
Only the forth expression has (-6) as root.

Answer 9

another option is to use the factor theorem

find f(-4)
f(-6)
f(1)

ther equation where all 3 are zero is your answer...

which is D