Question

Exercițiul 5 doar cele încercuite , va rog!!

Answer 1

Răspuns:

"spargi convenabil termenul din mijloc in suma algebrica de 2 termeni

cauti ca produsul coeficientilor lui x sa iti dea termenul liber

Explicație pas cu pas:

c) x²+2x+5x+10=x(x+2) +5(x+2) =(x+2) (x+5)

j) x²+7x+8x+56=x(x+7)+8(x+7)=(x+7)(x+8)

g) x²+3x+5x+15=x(x+3)+5(x+3)=(x+3)(x+5)

d) x²+2x+3x+6=x(x+2)+3(x+2)=(x+2) (x+3)

Answer 2

$\it c)\ x^2+7x+10=x^2+2x+5x+10=x(x+2)+5(x+2)=(x+2)(x+5)\\ \\ \\ d)\ x^2+5x+6=x^2+2x+3x+6=x(x+2)+3(x+2)=(x+2)(x+3)\\ \\ \\ g)\ x^2+8x+15=x^2+3x+5x+15=x(x+3)+5(x+3)=(x+3)(x+5)\\ \\ \\ j)\ x^2+15x+56=x^2+7x+8x+56=x(x+7)+8(x+7)=(x+7)(x+8)$

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$\it c)\ x^2+7x+10=x^2+7x+14-4=(x^2-4)+(7x+14)=\\ \\ =(x^2-2^2)+7(x+2)=(x-2)(x+2)+7(x+2)=(x+2)(x-2+7)=\\ \\ =(x+2)(x+5)\\ \\ \\ d)\ x^2+5x+6=x^2+5x+10-4=(x^2-4)+(5x+10)=\\ \\ =(x-2)(x+2)+5(x+2)=(x+2)(x-2+5)=(x+2)(x+3)$

$\it g)\ x^2+8x+15=x^2+8x+24-9=(x^2-9)+(8x+24)=\\ \\ =(x-3)(x+3)+8(x+3)=(x+3)(x-3+8)=(x+3)(x+5)$

$\it j)\ x^2+15x+56=x^2+14x+x+49+7=(x^2+14x+49)+(x+7)=\\ \\ =(x+7)^2+(x+7) =(x+7)(x+7)+(x+7)=(x+7)(x+7+1)=\\ \\ =(x+7)(x+8)$