Factoring Polynomials?

Question

x^2+11x+24



x^3+27y^3

Answer 1

x^2 + 11x + 24
What are two factors of 24 which add to 11?
{1 & 24 , 2 & 12 , 3 & 8 , 4 & 6}

x^2 + 3x + 8x +24 re-write 11 x
x(x + 3) + 8(x + 3) common factor from each pair
(x + 3)(x + 8) common factor of (x + 3)
-----------------------
x^3 + 27y^3 sum of 2 cubes

use the formula a^3 + b^3 = (a + b)(a^2 - ab + b^2)

x^3 + (3y)^3
(x + 3y) [ x^2 - (x)(3y) + (3y)^2 ]
(x + 3y)(x^2 - 3xy + 9y^2)

Answer 2

x^2 +11x +24 = (x+8)(x+3)

x^3+27y^3 = (x+3y)(x^2-3xy+9y^2)

Answer 3

x^2+11x+24
=x^2+3x+8x+24 -------(here basically we have to find two nos whose sum is 11 & whose product is 1 x 24 =24)
=x(x+3)+8(x+3)
=(x+3)(x+8) ans

x^3+27y^3
={(x)^3+(3y)^3}
=(x+3y){x^2- x.3y+(3y)^2} ---------using a^3+b^3=(a+b)(a^2-ab+b^2)
=(x+3y)(x^2-3xy+9y^2) ans

Answer 4

x² + 11x + 24 = 0

x² + 3x + 8x + 24 = 0

Group factor

(x² + 3x) + (8x + 24) = 0

x(x + 3) + 8(x + 3) = 0

(x + 8)(x + 3) = 0

- - - - - - -s-

Answer 5

(x+8)(x+3)


(x+3)(x^2 +3x + 9)

Answer 6

x^2+11x+24=(x+3)(x+8)
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x^3+27y^3=(x+3y)(x^2-3yx+9y^2)