Question

HELP x^3-7x^2-x+7=0 make a list of possible roots, write the resulting quadratic function, factor or use quadratic formula to find the final 2 roots

Answer 1

all possible roots, \[\pm1, \pm7\]
try plugging in the x, for eg x=-1, the equation goes to 0.
thus (x+1) is a factor. (x+1)=0, x=-1
then \[(x+1)(Ax ^{2}+Bx+7)=0\]
to find the corresponding quadratic equation after taking out (x+1),
compare both brackets and group x of the same powers for the A and B, and take it equals to the original equation of the same power

eg for B: Bx(1)+7(x)=-x
B+7=-1
B=-8

for A, \[(1)(Ax ^{2})+Bx(x)=-7x ^{2}\]
A+B=-7
A=1

there fore your equation will be \[(x+1)(x ^{2}-8x+7)=0\]
then solve your quadratic equation

Answer 2

do you understand the AXsq+BX+7 thingy there? i scare my explanation is not clear. but thats how i solve my question lol

Answer 3

can u help with this one also? x^4+x^3-6x^2-14x-12=0 mkae a list of possible rational roots, write the resulting cubic function , use synthetic division to find a second root that will reduce the cubic expression to a quadratic expression

Answer 4

apply the same method, the possible roots can be obtained from the possible factors of the constant of your function, means\[\pm1, \pm 2, \pm3, \pm4, \pm6, \pm12\]plug into the equation and see which will make the equation goes to 0, then u can obtain your 1st factor.
use division to reduce it.

Answer 5

hi do u get the trick?

Answer 6

hi do u get the trick?