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Q: How do you factor 3x squared minus 8x minus 3?

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3x2 - 8x = x(3x - 8)

(z - 8x)(z + 8x)

(3x - 8)(8x + 5)

(8x - 3)(8x + 3)

4X^2 - 8X factor out 4X 4X(X - 2)

8x2-3x-5 = (8x+5)(x-1) when factored

(2x - 1)(4x + 1)

(8x + 9)(3x^2 - 1)

8X^2 - 4X^2 factor out a 4X^2 4X^2(4 - 1) all I can see to do

There is a formula for the "difference of squares." In this case, the answer is (8x + 5y)(8x - 5y)

5x

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)

the answer of √8x+16-3x=0 is x= 0.447213595 = +/- 0.447(to 3 decimal places) if I'm not mistaken

(4x - 3)(2x + 3)

(x - 5)(x - 3)

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

8x

8x(1 - 3x)

What about it? Factorise it?

5x2 + 3x + 8x2 = 13x2 + 3x = x(13x + 3)

(4x-8)(2x+2)

x2 - 8x = x*x - 8*x = x*(x - 8)

They are like terms and can be added together as 11x squared

8x2 -18x+9=(2x+-3)(4x+-3)

The given expression can be simplified to: 8x squared +9