Math Question?

Question

x^3 - 7x + 6 = 0. Find the factorization of the polynomial. Show all work.

Answer 1

x^3-x-6x+6=0
x(x^2-1)-6(x-1)=0
x(x-1)(x+1)-6(x-1)=0
(x-1)(x(x+1)-6)=0
(x-1)(x^2+3x-2x-6)=0
(x-1)[x(x+3)-2(x+3)]=0
(x-1)(x+3)(x-2)=0 then x-1=0 or x-2=0 or x+3=0
x=1 or x=2 or x=-3

Answer 2

Let p(x) = x^3 - 7x + 6

What we need to do is find x such that p(x) = 0. The key numbers we need to test are the factors of the constant term, 6. Thus, we need to test: -1, 1, -2, 2, -3, 3

As soon as we have one of these values work, we'll know that x minus that number will be a factor and we can use long division.

Test p(-1):

p(-1) = -1 + 1 - 6, which is NON-zero.
p(1) = 1 - 7 + 6 = 7 - 7 = 0

Since 1 is a root, (x - 1) is a factor.

So now, we use long division now that we know (x - 1) is a factor.

x - 1 INTO x^3 - 7x + 6

Rather than try and type out the long division, I'll leave this task up to you and write the final solution here.

Your answer should be x^2 + x - 6 as a result of the division, so that now becomes the result of our factoring.

x^3 - 7x + 6 = 0
(x - 1) (x^2 + x - 6) = 0

Now we can factor the second set of brackets easily, as it's just a quadratic.

(x - 1) (x + 3) (x - 2) = 0

And now solving it is trivial. x = {1, -3, 2}

Answer 3

First you have to guess one of the solutions which from the looks of the equation isn't difficult. One of the solutions is 1, because 1^3 - 7 * 1 + 6 = 0.

Now we have to devide the existing polynom by x - 1.

(1) (x^3 - 7x + 6) / (x - 1) = x^2 .....

when we multiply x^2 by (x - 1) we get

(2) x^3 - x^2

Now we subtract (2) from original polynom and get the following reminder

(3) x^2 - 7x + 6

now we continue dividing (3) by x - 1

(4) (x^2 - 7x + 6) / (x - 1) = x ...

in the same manner we multiply x with x - 1 and subtract it from (3). We get the following remainder

(5) -6x + 6

now we devide (5) by x - 1

(6) (-6x + 6) / (x - 1) = -6

if we multiply -6 with x -1 and subtract it from (5) we get remainder of 0 which we should get since 1 is solution to the equation.

Now the result of deviding x^3 - 7x + 6 with x - 1 can be read from (1), (4) and (6) as

(7) x^2 + x - 6

now all we have to do is find the solutions to x^2 + x - 6 = 0 which is given by formula for quadratic equation solutions

x1 = (-b + sqrt(b^2 - 4ac))/2a, x2 = (-b - sqrtsqrt(b^2 - 4ac))/2a

So x1 = 2, x2 = -3

Therefore

x^3 - 7x + 6 = (x - 1)(x - 2)(x + 3)

Answer 4

let f(x)= x^3-7x+6

f(1) = 1-7+6
= 0.

So, x-1 is a factor of f(x)

by dividing x^3-7x+6/ x-1 , we get x^2+x-6.

so factors of f(x) are (x+1) , (x+3) and (x-2)

so, x= -1 or -3 or 2