a 2(x - 4)^2 + 10
b 2(x - 4)^2 - 6
c 2(x - 4)^2 - 22
d 2(x + 4)^2 + 10
e 2(x + 4)^2 - 6
f 2(x + 4)^2 -22
2007-09-17 07:19:03 · 12 answers · asked by Niall R 1 in Science & Mathematics ➔ Mathematics
2 (x² - 8x + 5)
2(x² - 8x + 16 - 16 + 5)
2(x² - 8x + 16 - 11)
2(x² - 8x + 16) - 22
2(x - 4)² - 22
Option c
2007-09-21 11:18:04 · answer #1 · answered by Como 7 · 1⤊ 1⤋
2 x^2-16 x+10
2007-09-24 23:10:24 · answer #2 · answered by Anonymous · 0⤊ 0⤋
c
because
2x² - 16x + 10 = 2(x² - 8x ) + 10
= 2(x² - 8x + 16) + 10 - 32
= 2(x - 4)² - 22
Steps:
1. take out a factor of 2 (the coefficient of x²)
2. to make a perfect square halve the coefficient of x and square it, add it inside the brackets and subtract 32 outside the brackets since you cannot change the original expression
3. the brackets will now contain a perfect square (always). Factorise it.
2007-09-17 14:37:44 · answer #3 · answered by Anonymous · 0⤊ 1⤋
= 2x^2 - 16x + 10
= 2(x - 16)^2 + 10
= 2(x - 4)^2 + 10 - 16
= 2(x - 4)^2 - 6
The answer is letter b, 2(x - 4)^2 - 6.
2007-09-23 03:25:24 · answer #4 · answered by Jun Agruda 7 · 2⤊ 0⤋
B)
(x-4)^2 = x^2 -8x +16
2(x^2 - 8x + 16) - 6 = 2x^2 - 16x +10
2007-09-17 14:32:24 · answer #5 · answered by Anonymous · 0⤊ 2⤋
2x^2 - 16x + 10 =
2(x^2 - 8x) + 10 =
2(x - 4)^2 + 10 - 16 =
2(x - 4)^2 - 6
B
2007-09-17 14:31:06 · answer #6 · answered by PMP 5 · 0⤊ 1⤋
2 (x-4)^2= -10
2(x-4)(x-4) = -10
2[ x^2-8x+16] = -10
2x^2-16x+32 =-10
I guess C is the correct answer
2007-09-24 19:21:41 · answer #7 · answered by Will 4 · 0⤊ 0⤋
b)
(x-4)^2 = x^2 -8x +16
2(x^2 - 8x + 16) - 6 = 2x^2 - 16x +10
2007-09-17 14:30:59 · answer #8 · answered by skipper 7 · 1⤊ 1⤋
c is the correct answer
multiplied out it gives the answer you are looking for
2007-09-17 14:33:13 · answer #9 · answered by Aslan 6 · 0⤊ 1⤋
I would go with E
2007-09-23 13:46:24 · answer #10 · answered by scide i 2 · 0⤊ 0⤋