jika (ax+b)(3x-4)=cx2+7x-20 tentukan a+b+c

jawab

(ax + b)(3x - 4)= cx^2 +7x - 20

ax(3x -4) + b(3x - 4) =  cx^2 + 7x - 20

3a x^2 - 4ax + 3b x - 4b = cx^2 +7x - 20
(3a) x^2 + (-4a+3b) x + (-4b)=  (c) x^2 + (7) x + (-20)
koefisien kiri = koefisien kanan

3a = c
-4b = -20
b = 5

-4a + 3b = 7
-4a + 15 = 7
-4a = - 8
a = 2

c = a = 2

a+ b + c = 2 + 5 + 2 = 9

jawab :

(ax + b)(3x - 4) = cx² + 7x -20

(3a)x² - (4a + 3b) x - (4b) = (c)x² + (7) - (20)

(koefisien x dihilangkan)

-4b = -20
b = 5

-4a + 3b = 7
-4a + 3(5) = 7
-4a = 7-15
-4a = -8
a = 2

3a = c
3(2) = c
6 = c

jadi,
a + b + c
= 2 + 5 + 6
= 13

semoga membantu.......