How to write 5(2x+1)(x+1) in standard form explain....?

2 answers:

6 0

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.

Yuki888 [10]3 years ago

4 0

Answer:

$=10x^2+15x+5$

Step-by-step explanation:

$5\left(2x+1\right)\left(x+1\right)\\\mathrm{Expand}\:\left(2x+1\right)\left(x+1\right):\quad 2x^2+3x+1\\\mathrm{Apply\:FOIL\:method}:\quad \left(a+b\right)\left(c+d\right)=ac+ad+bc+bd\\a=2x,\:b=1,\:c=x,\:d=1\\=2xx+2x\cdot \:1+1\cdot \:x+1\cdot \:\\=2xx+2\cdot \:1\cdot \:x+1\cdot \:x+1\cdot \:1\\\mathrm{Simplify}\:2xx+2\cdot \:1\cdot \:x+1\cdot \:x+1\cdot \:1:\quad 2x^2+3x+1\\2xx+2\cdot \:1\cdot \:x+1\cdot \:x+1\cdot \:1\\2xx=2x^2\\2\cdot \:1\cdot \:x=2x\\1\cdot \:x=x\\1\cdot \:1=1\\=2x^2+2x+x+1$

$\mathrm{Add\:similar\:elements:}\:2x+x=3x\\=2x^2+3x+1\\=5\left(2x^2+3x+1\right)\\\mathrm{Expand}\:5\left(2x^2+3x+1\right):\quad 10x^2+15x+5\\5\left(2x^2+3x+1\right)\\\mathrm{Distribute\:parentheses}\\=5\cdot \:2x^2+5\cdot \:3x+5\cdot \:1\\\mathrm{Simplify}\:5\cdot \:2x^2+5\cdot \:3x+5\cdot \:1:\quad 10x^2+15x+5\\5\cdot \:2x^2+5\cdot \:3x+5\cdot \:1\\\mathrm{Multiply\:the\:numbers:}\:5\cdot \:2=10\\=10x^2+5\cdot \:3x+5\cdot \:1\\\mathrm{Multiply\:the\:numbers:}\:5\cdot \:3=15$