Question

factorise (x+a)(x+b)(x-c)​

Answer 1

Answer:

When we have

(x-a)(x-b)(x-c) you should multiply the first two to get a quadratic then multiply the final factor.

(x-a)(x-b)(x-c)

= (X-C)(x^2 - (a+b)x + ab)

= x^3 - (à+b+c)x^2 + (ab+bc+ac)x + abc

notice that whatever happens to à happens to a b and c too.

(x-a)(x-b)(x-c)(x-d)

=(x-d)(x-c)(x^2 - (a+b)x + ab)

= x^4 + (a+b+c+d)x^3 + (ab + bc + bd + ac + ad +cd)x^2 + (abc + bcd + abd + acd)x + abcd

Now we notice the amount of factors is the binomial coefficient. 1 4 6 4 1 for 4 factors. For n factors it’s 1 n n(n-1)/2

Answer 2

Step-by-step explanation:

We know

x^3 + y^3 + z^3 - 3xyz = ( x + y + z ) [ x^2 + y^2 + z^2 - xy - yz - zx ]

So we apply this identity and get our given equation As :

⇒[ ( x - a ) + ( x - b ) + ( x - c ) ] [ ( x - a )^ 2 + ( x - b )^2 + ( x - c )^2 - ( x - a ) ( x - b ) - ( x - b )( x - c ) - ( x - c ) ( x - a ) ]

⇒[ ( x - a ) + ( x - b ) + ( x - c ) ] [ ( x - a ) { ( x - a ) - ( x - b ) } + ( x - b ) { ( x - b ) - ( x - c ) } + ( x - c ) { ( x - c ) - ( x - a ) } ]

⇒[ ( x - a ) + ( x - b ) + ( x - c ) ] [ ( x - a ) {( x - a - x + b ) } + ( x - b ) { ( x - b - x + c ) } + ( x - c ) { ( x - c - x + a ) } ]

⇒[ ( x - a ) + ( x - b ) + ( x - c ) ] [ ( x - a ) { b - a } + ( x - b ) { c - b } + ( x - c ) { a - c } ]

⇒[ ( x - a ) + ( x - b ) + ( x - c ) ] [ ( bx - ax - ab + a^2 ) + (cx - bx - bc + b^2 ) + ( ax - cx - ac + c^2 ) ]

⇒[ ( x - a ) + ( x - b ) + ( x - c ) ] [ bx - ax - ab + a^2 + cx - bx - bc + b2 + ax - cx - ac + c^2 ]

⇒[ ( x - a ) + ( x - b ) + ( x - c ) ] [ - ab + a^2 - bc + b^2 - ac + c^2 ]

⇒[ x - a + x - b + x - c ] [ a^2 + b^2 + c^2 - ab - bc - ac ]

⇒[ 3x - ( a + b + c ) ] [ a^2 + b^2 + c^2 - ab - bc - ac ]