Question

If a+b+c=6, a^2+b^2+c^2=26 find the value of ab+bc+ca​

Answer 1

Answer:

5

Step-by-step explanation:

a² + b² + c² = (a + b + c)² - 2 (ab + bc + ca)

26=6^2-2(ab+bc+ca)

26-36= -2(ab+bc+ca)

-10 = -2(ab+bc+ca)

5= ab+bc+ca

Answer 2

Step-by-step explanation:

Answer:

Here one root is x = 1 so (x - 1) is a factor. i.e

{x}^{3} - {x}^{2} - 3 {x}^{2} + 3x + 2x - 2 =x

3

−x

2

−3x

2

+3x+2x−2=

{x}^{2} (x - 1) - 3x(x - 1) + 2(x - 1) =x

2

(x−1)−3x(x−1)+2(x−1)=

(x - 1)( {x}^{2} - 3x + 2) =(x−1)(x

2

−3x+2)=

(x - 1)( {x}^{2} - 2x - x + 2) =(x−1)(x

2

−2x−x+2)=

(x - 1)(x(x - 2) - 1(x - 2)) =(x−1)(x(x−2)−1(x−2))=

(x - 1)(x - 2)(x - 1) = (x - 1) {}^{2} (x - 2)(x−1)(x−2)(x−1)=(x−1)

2

(x−2)