1.So the problem "y=2x+5" is in standard form and no modification is necessary. For a parabolic equation, the standard form is y = a(x - h)^2 + k, from which direction (polarity of "a") and axis of symmetry (value of "h"), etc.
2.Ax + By + C = 0 or Ax + By = C.
Answer:
[tex]=10x^2+15x+5[/tex]
Step-by-step explanation:
[tex]5\left(2x+1\right)\left(x+1\right)\\\mathrm{Expand}\:\left(2x+1\right)\left(x+1\right):\quad 2x^2+3x+1\\\mathrm{Apply\:FOIL\:method}:\quad \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd\\a=2x,\:b=1,\:c=x,\:d=1\\=2xx+2x\cdot \:1+1\cdot \:x+1\cdot \:\\=2xx+2\cdot \:1\cdot \:x+1\cdot \:x+1\cdot \:1\\\mathrm{Simplify}\:2xx+2\cdot \:1\cdot \:x+1\cdot \:x+1\cdot \:1:\quad 2x^2+3x+1\\2xx+2\cdot \:1\cdot \:x+1\cdot \:x+1\cdot \:1\\2xx=2x^2\\2\cdot \:1\cdot \:x=2x\\1\cdot \:x=x\\1\cdot \:1=1\\=2x^2+2x+x+1[/tex]
[tex]\mathrm{Add\:similar\:elements:}\:2x+x=3x\\=2x^2+3x+1\\=5\left(2x^2+3x+1\right)\\\mathrm{Expand}\:5\left(2x^2+3x+1\right):\quad 10x^2+15x+5\\5\left(2x^2+3x+1\right)\\\mathrm{Distribute\:parentheses}\\=5\cdot \:2x^2+5\cdot \:3x+5\cdot \:1\\\mathrm{Simplify}\:5\cdot \:2x^2+5\cdot \:3x+5\cdot \:1:\quad 10x^2+15x+5\\5\cdot \:2x^2+5\cdot \:3x+5\cdot \:1\\\mathrm{Multiply\:the\:numbers:}\:5\cdot \:2=10\\=10x^2+5\cdot \:3x+5\cdot \:1\\\mathrm{Multiply\:the\:numbers:}\:5\cdot \:3=15[/tex]
[tex]=10x^2+15x+5\cdot \:1\\\mathrm{Multiply\:the\:numbers:}\:5\cdot \:1=5\\=10x^2+15x+5\\[/tex]
