If x-2 and x+2 are factors of 6x^3 + ax^2 + bx + 16, find a, b & any other factors

[popcorn123]

If x-2 and x+2 are factors of 6x^3 + ax^2 + bx + 16, find a, b & any other factors

If x-2 and x+2 are factors of 6x^3 + ax^2 + bx + 16, determine the values of a and b and any remaining factors.

 

[stapel]

If x-2 and x+2 are factors of 6x^3 + ax^2 + bx + 16, determine the values of a and b and any remaining factors.

So... you did the division (or the evaluation), set the remainders equal to zero, and... then what?

Please be complete. Thank you!

 

[Prove It]

If x-2 and x+2 are factors of 6x^3 + ax^2 + bx + 16, determine the values of a and b and any remaining factors.

If (x - 2) is a factor, then P(2) = 0.

If (x + 2) is a factor, then P(-2) = 0.

This will give you two equations in two unknowns, you will be able to solve for a and b.

 

[stapel]

If (x - 2) is a factor, then P(2) = 0.

If (x + 2) is a factor, then P(-2) = 0.

Remember to define all variables (especially for students who seem to be struggling badly). In this case, is it safe to assume that, by "P(x)", you mean "P(x) = y = 6x^3 + ax^2 + bx + 16"?

Thank you!

 

[popcorn123]

If (x - 2) is a factor, then P(2) = 0.

If (x + 2) is a factor, then P(-2) = 0.

This will give you two equations in two unknowns, you will be able to solve for a and b.

I got P(2) = 4a + 2b = -64
P( -2) = 4A - 2B = 32

Solving for 2 equations, 2 unknown, I got a = -4 and b = -24.

To find the remaining factor, I did this:

6x^3 - 4x^2 - 24x + 16
2x^2 (3x - 2) - 8( 3x -2)
(3x - 2) (2x^2 - 8)


So, at this point (2x^2 - 8) can't factor more ... so the other root would be x= 2/3, right?

So what does the website say when it puts the solution as, "remaining factors = [FONT=MathJax_Main]2[FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])" mean? [/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])[/FONT] "[/FONT]

 

[Ishuda]

I got P(2) = 4a + 2b = -64
P( -2) = 4A - 2B = 32

Solving for 2 equations, 2 unknown, I got a = -4 and b = -24.

To find the remaining factor, I did this:

6x^3 - 4x^2 - 24x + 16
2x^2 (3x - 2) - 8( 3x -2)
(3x - 2) (2x^2 - 8)


So, at this point (2x^2 - 8) can't factor more ... so the other root would be x= 2/3, right?

So what does the website say when it puts the solution as, "remaining factors = [FONT=MathJax_Main]2[FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])" mean? [/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Math]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main])[/FONT] "[/FONT]

The red above is incorrect:
(2x^2 - 8) = 2 (x² - 4) = 2 (x - 2) (x + 2)
This has an extra 2 which must go with the remaining factor, thus the remaining factors are 2 and (3x-2) and not just (3x-2)

Also, yes the other root would be x=2/3.

 

[popcorn123]

The red above is incorrect:
(2x^2 - 8) = 2 (x² - 4) = 2 (x - 2) (x + 2)
This has an extra 2 which must go with the remaining factor, thus the remaining factors are 2 and (3x-2) and not just (3x-2)

Oh right, you can take out a 2. Okay this makes sense. Thank you,

 

[Prove It]

The red above is incorrect:
(2x^2 - 8) = 2 (x² - 4) = 2 (x - 2) (x + 2)
This has an extra 2 which must go with the remaining factor, thus the remaining factors are 2 and (3x-2) and not just (3x-2)

Oh right, you can take out a 2. Okay this makes sense. Thank you,

As it happens anyway, even if the numbers are not perfect squares, as long as there is a DIFFERENCE then your expression can still be factorised using DOTS, where \(\displaystyle a - b = \left( \sqrt{a} \right) ^2 - \left( \sqrt{b} \right) ^2 = \left( \sqrt{a} - \sqrt{b} \right) \left( \sqrt{a} + \sqrt{b} \right) \), so you should NEVER think that a difference is unable to be factorised.

A sum on the other hand is another story...