(6) \( x^{2}+4 x+k \) (7) \( 9 x^{2}+k x+16 \) (8) \( k x^{2}-20 x+4 \) Factorize of each of the following complet (9) \( x^{3}+12 x^{2}+36 x \) (10) \( 4 x^{2}-28 x+49 \)

(6) We wish to write \[ x^2+4x+k=(x+a)(x+b) \] so that when expanded \[ (x+a)(x+b)=x^2+(a+b)x+ab, \] we must have \[ a+b=4\quad\text{and}\quad ab=k. \] Thus the quadratic factors (over the integers, say) if and only if two numbers exist whose sum is 4 and whose product is \(k\). (For example, if one chooses \(a=2\) and \(b=2,\) then \(k=2\cdot2=4,\) so that \[ x^2+4x+4=(x+2)^2. \] Otherwise the answer is best left in the form \[ x^2+4x+k=(x+a)(x+b)\quad\text{with}\quad a+b=4\text{ and }ab=k. \] (7) Similarly, write \[ 9x^2+kx+16=(3x+p)(3x+q). \] Expanding the right‐side gives \[ (3x+p)(3x+q)=9x^2+3(p+q)x+pq. \] Thus we need \[ 3(p+q)=k\quad\text{and}\quad pq=16. \] That is, the expression factors completely (over the integers, if possible) once two numbers \(p\) and \(q\) are found satisfying \[ p+q=\frac{k}{3}\quad\text{and}\quad pq=16. \] For example, if we take \(p=q=4\) (since \(4\cdot4=16\)) then \(p+q=8\) and so \(k=3\cdot8=24\) and we obtain the perfect square factorization \[ 9x^2+24x+16=(3x+4)^2. \] (8) For \[ kx^2-20x+4, \] we wish to factor it completely. In many cases one may try to write it in the form \[ kx^2-20x+4=(ax+b)(cx+d), \] with \[ ac=k,\quad bd=4,\quad \text{and}\quad ad+bc=-20. \] A common approach is to “guess” numbers when \(k\) takes a convenient value. For instance, if we take \(k=24,\) then we have \[ 24x^2-20x+4. \] Notice that \[ 24x^2-20x+4= (4x-2)(6x-2). \] To check: \[ (4x-2)(6x-2)=24x^2-8x-12x+4=24x^2-20x+4. \] In general, unless further restrictions are given, one may express the factorization in “parameter‐form” by finding numbers \(a,b,c,d\) that satisfy \[ ac=k,\quad bd=4,\quad ad+bc=-20. \] (9) The cubic \[ x^3+12x^2+36x \] has a common factor of \(x.\) Factor this out: \[ x^3+12x^2+36x=x(x^2+12x+36). \] The quadratic factor is a perfect square because \[ x^2+12x+36=(x+6)^2. \] Hence the complete factorization is \[ x^3+12x^2+36x=x(x+6)^2. \] (10) Consider \[ 4x^2-28x+49. \] We can check that \((2x-7)^2\) expands to \[ (2x-7)^2=4x^2-28x+49. \] Thus the complete factorization is \[ 4x^2-28x+49=(2x-7)^2. \]