Question

factorise the equadratic trinomials :x*
2+10x+24​

Answer 1

Answer:

Step-by-step explanation:X²+10X+24=0   from the eqution, a=+1,b=+10,c=24

Method 1: Using factoisation by grouping

To factorise, we look a number when we multiply it gives "ac" and when when sumup those numbers it gives "b"

Factors of 24= 1,2,3,4,6,12,24   therefore the numbers are 6 and 4

⇒X²+6X+4X+24=0

⇒X(X+6)+4(X+6)=0

⇒(X+6)(X+4)=0

⇒X+6=0  or  X+4=0

⇒X= -6  or  X= -4

Method 2; Using formula method X= [-b ± √(b²-4ac)]/2a = (-b ± √Δ)/2a

Δ=b²-4ac

  =10²-4(1)(24)

  =100-96

Δ =4

X=(-b ± √Δ)/2a = (-10±√4)/2(1)

                    = (-10±2)/2

                     = (-10+2)/2    or   (-10-2)/2

                    = -8/2   or    -12/2

                  X  = -4  or  -6

 

Answer 2

QUESTION SHOULD BE:

• Factorise the quadrtic polynomial: x²+10x+24.

ANSWER:

• Factorisation of the above expression is (x+4)(x+6).

GIVEN:

• P(x) = x²+10x+24

TO FIND:

• Factorise the above expression.

SOLUTION:

= x²+10x+24

= x²+4x+6x+24

= (x²+4x)+(6x+24)

= x(x+4) +6(x+4)

= (x+4)(x+6)

Factorisation of the above expression is (x+4)(x+6).

NOTE:

Some important formulas:

(a+b)² = a²+b²+2ab

(a-b)² = a²+b²-2ab

(a+b)(a-b) = a²-b²

(a+b)³ = a³+b³+3ab(a+b)

(a-b)³ = a³-b³-3ab(a-b)

a³+b³ = (a+b)(a²+b²-ab)

a³-b³ = (a-b)(a²+b²+ab)

(a+b)² = (a-b)²+4ab

(a-b)² = (a+b)²-4ab