# Calculate the following expressions: a) (x+3)^2-(x-3)^2+(x+3)(x-3) b) (x^2+x+1)-(x+1)^2 c) (x+2)(x^2-2x+4)-(x-3)(x^2+3x+9)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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a) E(x)= (x+3)^2-(x-3)^2+(x+3)(x-3)

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

(x-3)^2=x^2-2*3*x+9

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

E(x)= x^2+2*3*x+9 – (x^2-2*3*x+9) + x^2-9

E(x)= x^2+2*3*x+9 – x^2+2*3*x-9 + x^2-9

After the process of reducing similar terms, the expression will become:

E(x)= x^2 + 12x -9

b) E(x)= (x^2+x+1)-(x+1)^2

(x+1)^2= x^2 +2x + 1

E(x)= (x^2+x+1)-( x^2 +2x + 1)

If we are substituting x^2 +2x + 1 with a letter, t, the expression will become

E(t)=t-t

E(t)=0

So, the expression is not depending on the variable.

c) E(x)= (x+2)(x^2-2x+4)-(x-3)(x^2+3x+9)

E(x)=x*(x^2-2x+4)+2*(x^2-2x+4)-x*(x^2+3x+9)+3*(x^2+3x+9)

E(x)= x^3-2x^2+4x+2x^2-4x+8- x^3-3x^2-9x+3x^2+9x+27

After the process of reducing similar terms, the expression will become:

E(x)=8+27

E(x)=35

This expression does not depend on any variable, also.

neela | High School Teacher | (Level 3) Valedictorian

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We use (i) (a+b)^2 =a^2+2ab+b^2 ,(ii)(a+b)(a-b)= a^2-b^2. Of course (a-b)^2 = a^2-2ab+b^2 which is a consequence of (i)if we consider the sign. a) To simplify (x+3)^2-(x-3)^2+(x+3)(x-3). (x^2+6x+9)-(x^2-6x+9)+(x^2-3^2) =(1-1+1)x^2+(6--6)x+(9-9+3^2)= x^2+12x+9 b) (x^2+x+1)-(x+1)^2 = (x^2+x+1) -(x^2+2x+1) = (1-1)x^2+(1-2)x+1-1 = -x c) (x+2){x^2-2x+4)- (x-3)(x^2+3x+9) =x{(x^2-2x+4) -(x^2+3x+9)} + {2(x^2-2x+4)--3(x^2+3x+9)} =x(-5x-5) +(2+3)x^2+(-4+9)x+(8+27) =-5x^2-5x+5x^2+5x+35 = 35