Please faCtor. Thank you for your help.
a^2 + 4a + 21
typo I think, not negative 21?
NO it's postive 21
Sorry, all you can do is use complex roots then.
Look for the roots of a^2 + 4 a + 21 = 0
a = [-4 +/- sqrt (16-84)]/2
a = [-4 +/- sqrt (-68) ]/2
a = [-2 +/- 2 i sqrt_17 ] / 2
a = -1 + i sqrt 17 or a = -1 + i sqrt 17
so
(a + 1 + i sqrt 17)(a + 1 - i sqrt 17)
Would the answer be no souliton?
There is no real number solution.
Thank you for your help.
To factor the quadratic expression a^2 + 4a + 21, we need to find two binomials that, when multiplied, will give us the original expression.
Step 1: Look for two numbers that multiply to give the constant term (21) and add up to give the coefficient of the linear term (4).
In this case, the coefficient of the linear term is 4, and the constant term is 21. We need to find two numbers that multiply to give 21 and add up to 4. Those two numbers are 3 and 7 because 3 * 7 = 21 and 3 + 7 = 10.
The expression can now be rewritten as follows:
a^2 + 3a + 7a + 21
Step 2: Group the terms in pairs and factor by grouping.
(a^2 + 3a) + (7a + 21)
Step 3: Factor out the greatest common factor (GCF) of each group.
a(a + 3) + 7(a + 3)
Step 4: Notice that both terms have a common factor of (a + 3).
(a + 3)(a + 7)
Therefore, the factored form of the quadratic expression a^2 + 4a + 21 is (a + 3)(a + 7).