For all x, 12x^2 - 40x - 32 = 4(x - 4)(3x + 2)
how do we obtain this answer?
2007-06-10 14:06:54 · 6 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Remove the GCF = 4
4(3x^2 - 10x - 8)
Now find factors of 3x^2 [ only 3x and x] and of the 8 [either 4 x 2 or 8 x1] and using FOIL, see which give you the correct middle term of - 10x
4(3x + 2)(x - 4) -12x + 2x = -10x in the middle.
2007-06-10 14:13:06 · answer #1 · answered by richardwptljc 6 · 1⤊ 0⤋
12x^2 - 40x - 32 = 4(x - 4)(3x + 2)
12x^2 - 40x - 32 = 4(3x^2 + 2x - 12x - 8)
12x^2 - 40x - 32 = 12x^2 - 40x - 32
Hence this shows that the equation satisfies ALL values of x.
2007-06-10 21:18:02 · answer #2 · answered by Kemmy 6 · 0⤊ 0⤋
12x^2 - 40x - 32
=4(3x^2-10x-8)
=4(3x+2)(x-4)#
2007-06-10 21:12:45 · answer #3 · answered by jackleynpoll 3 · 0⤊ 0⤋
12x^2 - 40x - 32 = 4(x - 4)(3x + 2)
4(3x^2 - 10x - 8) = 4(x - 4)(3x + 2)
4(x - 4)(3x + 2) = 4(x - 4)(3x + 2)
.
2007-06-10 21:15:17 · answer #4 · answered by Robert L 7 · 0⤊ 0⤋
12x^2-40x-32
=4(3x^2-10x-8)
=4(3x^2-12x+2x-8)
=4(x-4)(3x+2)
2007-06-10 21:20:36 · answer #5 · answered by fair 2 · 1⤊ 0⤋
since this is a quadratic polynomial, use the quadratic formula for the form a x^2 + b x + c.
x = b^2 + ( (b^2 -4ac) )^(1/2)
x = b^2 - ( (b^2 -4ac) )^ (1/2)
or do long division.
2007-06-10 21:19:36 · answer #6 · answered by Drew 1 · 0⤊ 0⤋