Answered

At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

Help me on number 17 please

Help Me On Number 17 Please class=

Sagot :

RSTalbut
First get all the x terms on the left, so you get:
x³-6x²+8x=0
Then I used long division of polynomials and trial and error to factorise it:
(x³-6x²+8x)÷(x-2)=(x²-4x)
So:
(x³-6x²+8x)=(x²-4x)(x-2)=x(x-4)(x-2)
Therefore, for this cubic to equal 0, x=0, 4, or 2. 
I hope this is clear enough to help! :)
AL2006
x³ + 8x  =  6x²

Subtract  6x²  from each side:

x³ - 6x² + 8x  =  0

Before you go any further, factor 'x' out of the left side
and then let's talk about it:

x (x² - 6x + 8)  =  0

This equation is true when  x = 0 , so that's one of the solutions.

The other two will emerge when you figure out what makes (x² - 6x + 8) = 0.

Can that be factored ?  How about  (x - 4) (x - 2) ?

Now we can say that we need  (x - 4) (x - 2) = 0 .
That's true when  x = 4  or x = 2 .

So the three values of  'x'  that make the original equation a
true statement are . . .

x = 0
x = 2
x = 4
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.