9. When we factorise x2+5x+6, then we get:
A. (x + 2) (x + 3)
B. (x-2)(x-3)
C. (x × 2) + (x × 3)
D. (x × 2) –

2 thoughts on “9. When we factorise x2+5x+6, then we get:<br />A. (x + 2) (x + 3)<br />B. (x-2)(x-3)<br />C. (x × 2) + (x × 3)<br />D. (x × 2) –”

1. Answer:

   given equation : x² + 5x + 6

   Let us try factorizing this polynomial using splitting the middle term method.

   Factoring polynomials by splitting the middle term:

   We need to find two numbers ‘a’ and ‘b’ such that a + b =5 and ab = 6.

   On solving this we obtain, a = 3 and b = 2

   Thus, the above expression can be written as:

   x² + 5x + 6

   = x² + 3x + 2x + 6

   = x(x + 3) + 2(x + 3)

   = (x + 3)(x + 2)

   Thus, x+3 and x+2 are the factors of the polynomial x2 + 5x + 6.

   Reply

2. ⠀☆ GIVEN⠀ POLYNOMIAL – x² + 5x + 6 :

   ⠀⠀⠀⠀

   [tex]:\implies\sf x^2+3x + 6 = 0 \\\\\\\star \sf{By \:Using\:Sum-Product \:Pattern\::}\\\\ \:\implies\sf x^2 + 2x + 3x + 6= 0 \\\\\\\star\sf{Finding\:out\:Common\:Terms\::}\\\\\\:\implies\sf x(x + 2) +3(x + 2) = 0\\\\\\\star\sf{Now,\:Rewrite\:in\:Factored\:term\::}\\\\\\:\implies\sf (x + 2)\; (x + 3) = 0\\\\\\:\implies{\underline{\boxed{\frak{\purple{(x+2)(x+3)}}}}}\;\bigstar [/tex]

   Therefore,

   ⠀⠀⠀⠀⠀

   ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm { Hence, \:The\:factors\:are\:\bf{(x+2)\:and \:(x+3)\: }}}}\\[/tex]

   ⠀⠀⠀⠀⠀⠀[tex]{\underline{ \mathrm { \:The\:\:\: Option\:A\:is\:correct }}}\\[/tex]

   ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

   Reply