Factor x^2+4x+16

Problem

x2+4*x+16

Solution

1. Identify the coefficients of the quadratic expression a*x2+b*x+c where a=1 b=4 and c=16

2. Check for perfect square trinomial patterns. A perfect square trinomial follows the form a2+2*a*b+b2=(a+b)2 Here, a2=x2 and b2=16 which would mean a=x and b=4 However, the middle term 2*a*b would be 2(x)(4) = 8x,w*h*i*c*h*d(o)*e*s(n)*o*t*m*a*t*c*hx$.

3. Calculate the discriminant D=b2−4*a*c to determine if the quadratic can be factored over the set of real numbers.

D=4−4*(1)*(16)

D=16−64

D=−48

4. Conclude that since the discriminant is negative (D<0, the quadratic has no real roots and cannot be factored into linear factors with real coefficients. It is considered an irreducible quadratic over the real numbers.

5. Complete the square if a different form is required, though this is not a factorization into linear terms.

x2+4*x+16=(x2+4*x+4)+12

x2+4*x+16=(x+2)2+12

Final Answer

x2+4*x+16=irreducible over real numbers

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