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Assuming the missing symbols are pluses, that factors to (x + 4)(x + 1)

Q: What is the factorization of x2 5x 4?

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x2-5x+4 = (x-1)(x-4) when factored

If you mean: x2+5x+4 = 0 then the solutions are x = -1 and x = -4

To solve for x: x2 = 5x + 2 x2 - 5x = 2 x2 - 5x + (5/2)2 = 2 + (5/2)2 (x - 5/2)2 = 8/4 + 25/4 x - 5/2 = ± √(33/4) x = (5 ± √33) / 2

-x2 + 5x - 4 = 0. Then, x2 - 5x + 4 = (x - 4)(x - 1) = 0; whence, x = 1 or 4. Checking, -1 + 5 - 4 = 0; and -16 + 20 - 4 = 0.

x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4

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x2-5x+4 = (x-1)(x-4) when factored

It is: (x2 + 5x) (x + 8)

If you mean: x2+5x+4 = 0 then the solutions are x = -1 and x = -4

x2-5x+4 = (x-1)(x-4) when factord

When factored it is: (x-9)(x+4)

To solve for x: x2 = 5x + 2 x2 - 5x = 2 x2 - 5x + (5/2)2 = 2 + (5/2)2 (x - 5/2)2 = 8/4 + 25/4 x - 5/2 = ± √(33/4) x = (5 ± √33) / 2

-x2 + 5x - 4 = 0. Then, x2 - 5x + 4 = (x - 4)(x - 1) = 0; whence, x = 1 or 4. Checking, -1 + 5 - 4 = 0; and -16 + 20 - 4 = 0.