4x^2 - 29x + 7
(if it cannot be factored indicate if it is prime)
2007-07-08 04:30:20 · 4 answers · asked by D-pig 4 in Science & Mathematics ➔ Mathematics
4x^2 - 29x + 7
= 4x^2 - 28x - x + 7
= 4x (x-7) - 1(x-7)
= (4x-1)(x-7) ...final Answer
2007-07-08 04:33:54 · answer #1 · answered by Nterprize 3 · 1⤊ 0⤋
There is a simple test to see if one can factor
this over the integers:
If the polynomial is ax²+bx+c, compute
b²-4ac. If it is a square, the polynomial can be
factored. Otherwise not.
Here b²-4ac = 841-(28*4) = 729
and 729 = 27². So we can factor it and the factors are
(4x-1)(x-7).
2007-07-08 04:55:44 · answer #2 · answered by steiner1745 7 · 1⤊ 0⤋
(4x - 1)(x - 7)
2007-07-08 04:34:34 · answer #3 · answered by Anonymous · 1⤊ 0⤋
a million) 3x^3 - 9x^2 = 3(x^3 - 3x^2) = 3x^2(x - 3) (answer B) 2) 3y - 6x + 4xy - 2y^2 = 4xy - 2y^2 - 6x + 3y = (4xy - 2y^2) - (6x - 3y) = 2y(2x - y) - 3(2x - y) = (2x - y)(2y - 3) (answer C) 3) 12a^2 - 4ac + 3ab - bc = (12a^2 - 4ac) + (3ab - bc) = 4a(3a - c) + b(3a - c) = (3a - c)(4a + b) (answer D) 4) 35g^2h - 15gh^2 = 5(7g^2h - 3gh^2) = 5gh(7g - 3h) (answer A) 5) 6uv - 3v^3 + 4u - 2v^2 = -3v^3 - 2v^2 + 6uv + 4u = (-3v^3 - 2v^2) + (6uv + 4u) = -v^2(3v + 2) + 2u(3v + 2) = (3v + 2)(-v^2 + 2u) (answer A)
2016-10-20 07:05:41 · answer #4 · answered by Anonymous · 0⤊ 0⤋