What is the standard form of #y= (x+3) (x+4) #?

3 Answers
Jun 21, 2018

#y=x^2+7x+12#

Explanation:

the standard form of a second-degree polynomial function should look like this

#color(green)(y=ax^2+bx+c#

so we try to reach that for with your function

#y=(x+3)(x+4)#

Using the Distributive property

#y=x(x+4)+3(x+4)#

#y=x*x+x*4+3*x+3*4#

Multiply

#y=x^2+4x+3x+12#

Add like terms

#y=x^2+7x+12#

Jun 21, 2018

#x^2 + 7x + 12#

Explanation:

#y = (x+3)(x+4)#

The standard form of a quadratic equation is #ax^2 + bx + c#. To make this equation in standard form, we expand/simplify this using FOIL:
enter image source here

Following this image, we can multiply it out.

The #color(teal)("firsts")#:
#color(teal)(x*x) = x^2#

The #color(indigo)("outers")#:
#color(indigo)(x * 4) = 4x#

The #color(peru)"inners"#:
#color(peru)(3 * x) = 3x#

The #color(olivedrab)"lasts"#:
#color(olivedrab)(3 * 4) = 12#

Combine them all together to get:
#x^2 + 4x + 3x + 12#

We can still combine the like terms #color(blue)(4x)# and #color(blue)(3x)#:
#x^2 + 7x + 12#

As you can see, this matches the form #ax^2 + bx + c#. Therefore, the quadratic equation in standard form is #x^2 + 7x + 12#.

Hope this helps!

Jun 21, 2018

#y=x^2+7x+12#

Explanation:

#"the standard form of a quadratic is "#

#•color(white)(x)y=ax^2+bx+c color(white)(x);a!=0#

#"expanding the factors gives"#

#y=x^2+7x+12larrcolor(red)"in standard form"#