(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) $$ 4 x^{2}-28 x+49 $$

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(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
$$
4x^2-28x+49=(2x-7)^2.
$$
(6) $$ x^2 + 4x + k = (x + a)(x + b) $$ where $$ a + b = 4 $$ and $$ ab = k $$. (7) $$ 9x^2 + kx + 16 = (3x + p)(3x + q) $$ where $$ p + q = \frac{k}{3} $$ and $$ pq = 16 $$. (8) $$ kx^2 - 20x + 4 = (ax + b)(cx + d) $$ with $$ ac = k $$, $$ bd = 4 $$, and $$ ad + bc = -20 $$. (9) $$ x^3 + 12x^2 + 36x = x(x + 6)^2 $$. (10) $$ 4x^2 - 28x + 49 = (2x - 7)^2 $$.

(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) \( 4 x^{2}-28 x+49 \)

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