Matt's tangent

Let $p(x) = x^{4}-2x^{3}-9x^{2}+2x+8$ and $q(x) = ax+b$ . By sketching the graph of $p$ , we can easily see that the only way that $q$ can be tangent to $p$ in two distinct points is if $q$ is right at the bottom of $p$ , touching each 2 down-humps in a different point, let us say, $c$ and $d$ .

Now let us sketch $r(x) = p(x) - q(x)$ . Because $q$ is right at the bottom of $p$ , the difference $p(x) - q(x)$ is always positive, except for $x = c$ and $x = d$ , which make it go to zero. Therefore, $r$ is a polynomial with only 2 real roots $c$ and $d$ . Furthermore, $r$ is always positive when we get closer to $c$ and to $d$ , which means both have multiplicity 2 or greater.

Algebraically, $r(x) = x^{4}-2x^{3}-9x^{2}+(2-a)x+(8-b)$ and, by the conditions above, the only way to write $r$ in terms of $c$ and $d$ is $r(x) = (x-c)^{2}(x-d)^{2}$ .

From $x^{4}-2x^{3}-9x^{2}+(2-a)x+(8-b) = (x-c)^{2}(x-d)^{2}$ , we get $c+d = 1$ , $cd = -5$ , $a = -8$ and $b = -17$ . The answer is thus $|a+b| = 25$ .

The line $y=ax+b$ is tangent to the graph of $y=x^4-2x^3-9x^2+2x+8$ at exactly two distinct points if and only if the graph of $y=x^4 - 2x^3 -9x^2+(2-a)x+(8-b)$ touches the $x$ -axis at exactly two points.

This happens when $y=x^4 - 2x^3 -9x^2+(2-a)x+(8-b)$ has two repeated roots, or $x^4-2x^3-9x^2+(2-a)x+(8-b) = (x^2+cx+d)^2$ , for some real numbers $c$ and $d$ .

Expanding $(x^2+cx+d)^2$ , we get $x^4 +2cx^3 + (c^2+2d)x^2 +2cdx +d^2$ . By comparing coefficients, we get $2c=-2$ and $c^2+2d=-9$ , implying $c=-1$ and $d=-5$ . Thus, continuing to compare coefficients of the last two terms, we get $2-a=2cd=10$ and $8-b=d^2=25$ , so $a=-8$ and $b=-17$ .

Therefore $|a+b| = |-8-17|=\fbox{25}$

Subtract the two expressions to get $(x^4-2x^3-9x^2+2x+8) - (ax+b)$ .

Due to the tangency condition (the graph is now tangent to the x-axis at two distinct points), this expression has 2 double roots, e.g. $x^4-2x^3-9x^2+(2-a)x+(8-b) = (x-r_1)^2(x-r_2)^2$

Expanding the right hand side gives $(x-r_1)^2(x-r_2)^2\\=(x^2-2r_1x+r_1^2)(x^2-2r_2x+r_2^2) \\ = x^4-2(r_1+r_2)x^3+(r_1^2+4r_1r_2+r_2^2)x^2\\-2r_1r_2(r_1+r_2)x+(r_1r_2)^2$ Equating coefficients gives us the following system.

$\begin{cases} -2(r_1+r_2)=-2 \\ r_1^2+4r_1r_2+r_2^2=-9 \\ -2r_1r_2(r_1+r_2)=2-a \\ (r_1r_2)^2=8-b \end{cases}$ The first equation gives $r_1+r_2=1$ Subtracting the square of this from the second equation gives $(r_1^2+4r_1r_2+r_2^2)-(r_1^2+2r_1r_2+r_2^2)=-9 - 1^2\\ 2r_1r_2=-10\\ r_1r_2=-5$

Plugging these into the final 2 equations gives us values for $a, b$ . $-2(-5)(1)=2-a\\ a=-8$

$(-5)^2=8-b\\ b=-17$

Finally, we have our answer. $|a+b|=|-25|=\boxed{25}$

Let $y_1 = x^4 - 2 x^3 - 9 x^2 + 2 x + 8,$ and let $y_2 = a x+b$ be the line that touches the polynomial $y_1(x)$ at points $x_1$ and $x_2$ .

Then the curve described by the polynomial $y_3 \equiv y_1 - y_2 = x^4 - 2 x^3 - 9 x^2 + (2-a) x + 8 - b$ must touch the x-axis at the same points where the line $y_2(x)$ touches the polynomial $y_1(x)$ , that is, at $x_1$ and $x_2$ .

This means that each of the points $x_1$ and $x_2$ are double roots of $y_3(x)$ . This implies that we can write $y_3(x) = (x-x_1)^2 (x-x_2)^2$ Expanding the expression, $y_3(x) = x^4 -2(x_1+x_2) x^3 + [(x_1+x_2)^2 + 2 x_1 x_2] x^2 - 2 x_1 x_2 (x_1 + x_2) x + (x_1 x_2)^2,$ and comparing the two equations, we get $-2 = - 2 (x_1+x_2) \quad \Rightarrow \quad x_1+x_2 = 1;$ $(x_1+x_2)^2 + 2 x_1 x_2 = - 9 \quad \Rightarrow \quad x_1 x_2 = -5$ $2 - a = - 2 x_1 x_2 (x_1 + x_2) = - 2 \times 5 \times 1 = 10 \quad \Rightarrow \quad a = - 8$ $8 - b = (x_1 x_2)^2 = (-5)^2 = 25 \quad \Rightarrow \quad b = -17$ Therefore, $|a+b| = 25.$