Question

Consider the polynomial $p\left(x\right)=x^2+6x+k$.
(a) Show that, if $k=10$, this polynomial has no first degree factors.
(b) Give any negative value for k. write the resulting polynomial as a product of two first degree polynomials.
(d) Show that for any negative value of $k,p\left(x\right)$ has two distinct first degree factors.

Answer 1

$p\left(x\right)=x^2+6x+k$
(a) If k = 10
$p\left(x\right)=x^2+6x+10$
Discriminant $=b^2-4ac=6^2-4\left(1\right)\left(10\right)=36-40=-4$
Since the discriminant is negative, this equation has no solution. Hence, p(x) has no first degree factors.

(b) $p\left(x\right)=x^2+6x+k$
For p(x) to have first degree factor, the discriminant of the equation $x^2+6x+k=0$
should be zero or positive.
i.e. $b^2-4ac=6^2-4k\ge 0$
i.e. $6^2\ge 4k$
i.e. $\frac{36}{4}\ge k$
i.e. $k\le 9$
Thus, the maximum value of k is 9.

(c) If k = -1, then p(x) $=x^2+6x-1$
Consider $x^2+6x-1=0$
Discriminant $=b^2-4ac=6^2-4\left(1\right)(-1)=36+4=40$
Since the discriminant is positive, p(x) can be written as the product of two first degree polynomials.
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-6\pm 40}{2}$

$=\frac{-6+\sqrt{40}}{2},\frac{-6-\sqrt{40}}{2}$

$=\frac{-6+2\sqrt{10}}{2},\frac{-6-2\sqrt{10}}{2}$

$=-3+\sqrt{10},-3-\sqrt{10}$

$\therefore x^2+6x-1=(x-(-3+\sqrt{10}\left)\right)(x-(-3-\sqrt{10}\left)\right)$

$=(x+3-\sqrt{10})(x+3+\sqrt{10})$

(d) Let $k=-m\Rightarrow p\left(x\right)=x^2+6x-m$

In the second degree equation $x^2+6x-m=0$

Discriminant $=b^2-4ac=6^2-4\left(1\right)(-m)=36+4m$

Since this is positive, p(x) has two distinct first degree factors.