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Luv2Teach
You can do this using synthetic division, which is the easiest way.  If x - 2 = 0, then x = 2.  That 2 will go outside the "box" and the leading coefficients of the terms in the polynomial will go inside the "box".  2    (1   -3    -10    24).  Bring down the first number, the 1.  Multiply that 1 by the 2 to get 2.  Put that 2 up under the -3 and add to get -1.  Multiply that -1 by the 2 to get -2.  Put that =-2 up under the -10 and add to get -12.  Multiply that -12 by the 2 to get -24.  Put the -24 up under the 24 and add to get 0.  That means that x - 2 is a factor of the polynomial.  What's left, the bolded numbers, are the coefficients of a new polynomial that is one degree less than the polynomial you started with.  In other words, when we divide your polynomial by x-2, you get [tex]x^2-x-12[/tex].
xero099

[tex]\frac{x^{3}-3\cdot x^{2}-10\cdot x + 24}{x-2}[/tex] is equal to [tex]x^{2}-x -12[/tex].

In this question we check if [tex]x = 2[/tex] represents a root of the polynomial. If it does, then we apply algebra operations to simplify the expression:

1) Check if [tex]x = 2[/tex] is a root of the polynomial:

[tex]f(2) = (2)^{3} - 3\cdot (2)^{2} - 10\cdot (2) + 24[/tex]

[tex]f(2) = 0[/tex]

[tex]x = 2[/tex] is a root of the polynomial.

2) Algebra operations:

(i) [tex]\frac{x^{3}-3\cdot x^{2}-10\cdot x + 24}{x-2}[/tex]  Given.

(ii) [tex]\frac{(x+3)\cdot (x-4)\cdot (x-2)}{x-2}[/tex] Factoring.

(iii) [tex](x+3) \cdot (x-4)[/tex] Definition of division/Existence of multiplicative inverse/Modulative property

(iv) [tex]x^{2}-x -12[/tex] Distributive property/Commutative property/Associative property/[tex](-1)\cdot a = -a[/tex]/Result

We kindly invite to check this question on factoring: https://brainly.com/question/9781037

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