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# x2+3x+5 be factored explain why or why not

can you show the steps

### 1 Answer by Expert Tutors

Michael B. | Seasoned and experienced tutor with extensive science backgroundSeasoned and experienced tutor with exte...
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x2 + 3x + 5

For equations of the form:

ax2 + bx + c

where a = 1 it will factor to

(x + c1)(x + c2)

If you multiply that out you'll end up with

x2 + c1x + c2x + c1*c2 = x2 + (c1 + c2)x + c1*c2

This means that

b = c1+c2 and is the sum of the factors of c

c = c1*c2 and is the product of the factors of c

a = 1, b = 3, and c = 5

The only factors of a prime number are 1 and the number so c = 1*5

In order to factor it b has to equal the sum of those two numbers so it would be 6x not 3x.

This is why your equation cannot be factored in the standard "normal" way.

x = [-b ± √(b2 - 4ac)]/2a

b2 - 4ac = 9 - 20 = -11

so you have complex roots and

x = (-3+ i√11)/2

x = (-3 - i√11)/2

In factored form that would look like

(x + 3/2 - i√11/2)(x + 3/2 + i√11/2)

Multiplying that out to check you'll find

x2 + (3/2)x + (i√11/2)x + (3/2)x + 9/4 + (3i√11)/4 - (i√11/2)x(3i√11)/4 - i2(√11)2/4

Since i2 = -1 and (√11)2 = 11, the term on the end simplifies to 11/4