multiplying [7x + (y - 4)] x [7x - (y - 4)]

[zhyia]

i tried to solve [(7x+(y-4)] x [(7x-(y-4)]
but i came up with three different answers each time

(7x)x(7x)=49x^2
the positive and the neg y should cancel out
and that leaves the (-4)x(-4)=16

i came up with 49x^(16)^2
and 49x^2-(y+16)2
and 49x^2(y+16)^2
im not sure which is right?

 

[soroban]

Re: multiplying

Hello, zhyia!

Sorry, all your answers are wrong . . .

i tried to solve [7x+(y-4)][7x-(y-4)]

(7x)(7x) = 49x²

the positive and the neg y should cancel out . . . you're ADDING them?

There are at least two ways to simplify this one.


[1] We have: \(\displaystyle \,(7x\,+\,y\,-\,4)(7x\,-\,y\,+\,4)\)

Then multiply the two trinomials:

\(\displaystyle \;\;(7x)(7x)\,+\,(7x)(-y)\,+\,(7x)(4)\,+\,(y)(7x)\,+\,(y)(-y)\,+\,(y)(4)\,+\,(-4)(7x)\,+\,(-4)(-y)\ +\ (-4)(4)\)

\(\displaystyle = \;49x^2\,-\,\sout{7xy}\,+\,\sout{28x}\,+\,\sout{7xy}\,-\,y^2\,+\,4y\,-\,\sout{28x}\,+\,4y\,-\,16\)

\(\displaystyle = \;49x^2\,-\,y^2\,+\,8y\,-\,16\)


[2] Note that we have the form: \(\displaystyle \,(a\,+\,b)(a\,-\,b)\,\) which equals \(\displaystyle \,a^2\,-\,b^2\)

Hence: \(\displaystyle \,[7x\,+\,(y-4)][7x\,-\,(y-4)] \;=\;(7x)^2\,-\,(y-4)^2\) . . . etc.

 

[zhyia]

so my final answer should be 49x^2-y^2+8y-16 is that correct?