16y^2 -81
2007-03-09 03:15:09 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
16y^2 - 81
=(4y)^ 2- 9^2
=(4y - 9) (4y + 9)
2007-03-09 03:20:34 · answer #1 · answered by Malfoy vs Potter 5 · 0⤊ 0⤋
Others have given you the answer, so here is how to think about it.
Look at the first number. It looks like a perfect square (the result of multiplying a number by itself). Write that down
(4y- ) (4y+ )
Look at the last number. Same deal. Do the square root of that.
(4y-9)(4y+9) = 16y^2-81
Why is one minus and the other plus? Because the last number is negative. This can only happen when you multiply a positive with a negative.
2007-03-09 11:36:34 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Use the rule (a+b)(a-b)=a²-b²
So, 16y²-81=(4y+9)(4y-9).
You can always check your answer by expanding again
16y²-36y+36y-81
As -36y+36y=0, you are left with 16y²-81, therefore, the factorised form is correct.
2007-03-09 11:27:49 · answer #3 · answered by Forsaken 2 · 0⤊ 0⤋
(4y + 9)x(4y - 9) = 16y^2 - 81
2007-03-09 11:19:00 · answer #4 · answered by tuoidabuon 2 · 0⤊ 0⤋
(4y - 9)(4y + 9)
2007-03-09 11:20:34 · answer #5 · answered by aeiou 7 · 0⤊ 0⤋
= (4y - 9).(4y + 9)
2007-03-09 11:19:47 · answer #6 · answered by Como 7 · 0⤊ 0⤋