(х-2)*(х²+8х+16)=7*(х+4)

To solve the equation $(x-2)(x^2+8x+16)=7(x+4)$, we will simplify the expressions, factor where possible, and find the roots. 1. Simplify the Quadratic Expression Notice that $x^2+8x+16$ is a perfect square trinomial. It follows the pattern $a^2+2ab+b^2=(a+b{)}^2$: $x^2+8x+16=(x+4{)}^2$Substituting this back into the equation: $(x-2)(x+4{)}^2=7(x+4)$2. Move all terms to one side To avoid losing potential solutions by dividing by $(x+4)$, we move everything to the left side to set the equation to zero: $(x-2)(x+4{)}^2-7(x+4)=0$3. Factor out the common term Both terms share a factor of $(x+4)$. Let's factor it out: $(x+4)\cdot \left[\right(x-2\left)\right(x+4)-7]=0$4. Simplify the expression inside the brackets Expand the product $(x-2)(x+4)$: $(x-2)(x+4)=x^2+4x-2x-8=x^2+2x-8$Now substitute this back into the brackets: $(x+4)(x^2+2x-8-7)=0$ $(x+4)(x^2+2x-15)=0$5. Find the roots An equation is equal to zero if any of its factors are zero. Factor 1: $x+4=0⟹{\mathbf{x}}_{\mathbf{1}}\mathbf{=}-4$Factor 2: $x^2+2x-15=0$We can solve this quadratic equation using the quadratic formula or by factoring. Seeking two numbers that multiply to $-15$ and add to $2$, we find $5$ and $-3$: $(x+5)(x-3)=0$This gives us the remaining roots: $x+5=0⟹{\mathbf{x}}_{\mathbf{2}}\mathbf{=}-5$ $x-3=0⟹{\mathbf{x}}_{\mathbf{3}}\mathbf{=}\mathbf{3}$Answer: The solutions to the equation are $x=-5,-4,3$. Would you like me to verify these solutions by substituting them back into the original equation?