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x2 + 100 has no real solutions. Applying the quadratic formula, we find two imaginary solutions: Zero plus or minus 10itimes the square root of 1.

x = 10i

x = -10i

where i is the square root of negative 1.

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That doesn't factor neatly. Applying the quadratic formula, we find two imaginary solutions: (-1 plus or minus the square root of -31) divided by 2.

x = -0.5 + 2.7838821814150108i

x = -0.5 - 2.7838821814150108i

where i is the square root of negative one.

(a + b)2 - 100 = (a + b + 10)(a + b - 10).

x2 + 150 cannot be factored.

x=50

Q: How do you factor a plus b quantity squared minus 100?

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6(ab - ac + b2 - bc)

6ab(2x2+x-5)

(x - 5)(x - 3)

Remember to factor out the GCF of the coefficients if there is one. A perfect square binomial will always follow the pattern a squared plus or minus 2ab plus b squared. If it's plus 2ab, that factors to (a + b)(a + b) If it's minus 2ab, that factors to (a - b)(a - b)

(6x - 1)(6x - 1)

Related questions

This expression factors as x -1 quantity squared.

2m^2 - 8 -First you should factor out a two. --> 2(m^2-4) -You now have something squared minus something else squared; You have m squared minus 2 squared. Whenever you have something squared minus something squared as you do in this case, there is a simple rule to remember: You can reduce that expression into the quantity of the square root of the first number or variable plus the square root of the second number or variable Times the quantity of the square root of the first number or variable minus the second number or variable squared. --> In the case of your expression: ----> 2(m+2)(m-2)<-----

x squared minus 12 x plus 36 = (x - 6)(x - 6)

There is no rational factorisation.

There are no rational factors.

(9 - y)(a + 5)

(a - 2)(a^2 + 6)

(b-c)(a+b)-ac

(3y - 5)(y + 5)

(x + 3)(3x - 2)

68.3

(7a - 1)(6a + 1)