factor the trinomial ; r^3-2r^2-24r?

r^3-2r^2-24r
= r ( r^2 - 2r - 24)
= r ( r-6) (r+4)

sub x=-3 into equation

3(-3)^2 - 4(-3) +7
= 3(9) + 12 + 7
= 27 + 12 +7
= 46

1)
r^3 - 2r^2 - 24r
= r(r^2 - 2r - 24)
= r(r^2 + 4r - 6r - 24)
= r[(r^2 + 4r) - (6r + 24)]
= r[r(r + 4) - 6(r + 4)]
= r(r + 4)(r - 6)

2)
3x^2 - 4x + 7
= 3(-3)^2 - 4(-3) + 7
= 3(9) - (-12) + 7
= 27 + 12 + 7
= 46

r^3-2r^2-24r = r(r^2 -- 2r -- 24) = r(r -- 6)(r + 4)
3x^2 - 4x +7 = 3(--3)^2 -- 4(--3) + 7 = 27 +12 + 7 = 46

(-6 + r) r (4 + r)

Question 1
r ( r ² - 2 r - 24 )
r ( r - 6 ) ( r + 4 )

Question 2
27 + 36 + 7 = 70

r (r^2 - 2r -24) factor the r
r (r - 6)(r + 4) factor the trinomial
Answer is r (r - 6)(r + 4)

to evaluate just substitute -3 where ever x is and then solve the problem.

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