Q: How to factor 3x2 plus 8x plus 5?

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x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2

If you mean: 3x2+8+5 = 0 Then it crosses the x axis at points -1 and -5/3

Assuming the 2 is meant to be a square this is the form: x2 - 3x2 = -2x2 + 8x + 5

x2 + 8x + 15 = (x + 3) (x + 5).

If you notice, 8x + 4 = 4(2x + 1), but there is an odd coefficient for x2, so there is guaranteed to be some remainder (that is 8x + 4 is not a factor of the polynomial): (24x3 - 5x2 - 48x - 8) ÷ (8x + 4) = 3x2 - 2x - 5 - (x2 - 12)/(8x + 4)

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x2+8x = -3x2+5 4x2+8x-5 = 0 (2x+5)(2x-1) = 0 x = -5/2 or x = 1/2

5 - 8x + 3x2 = 3x2 - 8x + 5 (since the sign of the second term is negative, then 5 can be factored as (-1)(-5)) = (3x - )(x - ) (try to put 1 or 5 to the empty places, in order to obtain -8x) = (3x - 5)(x - 1)

= (3x + 5)(x + 1) so x = -5/3 or -1

If you mean: 3x2+8+5 = 0 Then it crosses the x axis at points -1 and -5/3

Assuming the 2 is meant to be a square this is the form: x2 - 3x2 = -2x2 + 8x + 5

x2 + 8x + 15 = (x + 3)(x + 5)

x2 + 8x + 15 = (x + 3) (x + 5).

If you notice, 8x + 4 = 4(2x + 1), but there is an odd coefficient for x2, so there is guaranteed to be some remainder (that is 8x + 4 is not a factor of the polynomial): (24x3 - 5x2 - 48x - 8) ÷ (8x + 4) = 3x2 - 2x - 5 - (x2 - 12)/(8x + 4)

3x2+13x-10 = (3x-2)(x+5) when factored

Twice. Between negative two and negative one.

(3x - 5)(x + 4)

(2x - 1)(2x + 5)