Question

15. Descompuneți în produs de factori:
a) (x²+x)(x²+x-8) + 12;
c) x(x + 4)(x² + 4x+6) + 8;
e) (x² + 3x)(x² + 3x + 9) + 14;
g) x(2x + 1)(2x² + x + 10) + 16;
i) (x² + 3x + 1)(x² + 3x-6) + 12;
b) x(x + 2)(x² + 2x - 14) + 48;
d) (x² + 4x + 1)(x² + 4x-2)- 18:
f) x(x + 2)(x² + 2x - 5) +6;
(x²-x)(x²-x-8) + 12;
j) (x²-5x)(x²-5x +11) + 30.

Answer 1

Răspuns:

Explicație pas cu pas:

a) (x²+x)(x²+x-8) + 12 = a(a-8)+12 = a²-8a+12 = a²-6a-2a+12 =

= a(a-6)-2(a-6) = (a-6)(a-2) = (x²+x-6)(x²+x-2)

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b) x(x + 2)(x² + 2x - 14) + 48 = b(b-14)+48 = b²-14b+48 =

= b²-8b-6b+48 = b(b-8)-6(b-8) = (b-8)(b-6) =

= (x²+2x-8)(x²+2x-6)

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c) c) x(x + 4)(x² + 4x+6) + 8 = c(c+6)+8 = c²+6c+8 = c²+4c+2c+8 =

= c(c+4)+2(c+4) = (c+4)(c+2) = (x²+4x+4)(x²+4x+2) = (x+2)²(x²+4x+2)

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d) (x² + 4x + 1)(x² + 4x-2)- 18 = (a+1)(a-2)-18 = a²-2a+a-2-18 =

= a²-a-20 = a²-5a+4a-20 = a(a-5)+4(a-5) = (a-5)(a+4) =

= (x²+4x-5)(x²+4x+4) = (x²+4x-5)(x+2)² = (x+5)(x-1)(x+2)²

x²+4x-5 = x²+5x-x-5 = x(x+5)-(x+5) = (x+5)(x-1)

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e) (x² + 3x)(x² + 3x + 9) + 14 = a(a+9)+14 = a²+9a+14 =

= a²+7a+2a+14 = a(a+7)+2(a+7) = (a+7)(a+2) =

= (x²+3x+7)(x²+3x+2) = (x²+3x+7)(x+1)(x+2)

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f) x(x + 2)(x² + 2x - 5) +6 = a(a-5)+6 = a²-5a+6 =

= a²-3a-2a+6 = a(a-3)-2(a-3) = (a-3)(a-2) =

= (x²+2x-3)(x²+2x-2)

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g) x(2x + 1)(2x² + x + 10) + 16 = a(a+10)+16 = a²+10a+16 =

= a²+8a+2a+16 = a(a+8)+2(a+8) = (a+8)(a+2) =

= (2x²+x+8)(2x²+x+2)

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h) (x²-x)(x²-x-8) + 12 = a(a-8)+12 = a²-8a+12 =

= a²-6a-2a+12 = a(a-6)-2(a-6) = (a-6)(a-2) =

= (x²-x-6)(x²-x-2)

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j) (x²-5x)(x²-5x +11) + 30 = a(a+11)+30 = a²+11a+30 =

= a²+6a+5a+30 = a(a+6)+5(a+6) = (a+6)(a+5) =

= (x²-5x+6)(x²-5x+5)

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