An algebra problem by Jade Mijares

The zeroes of $x^2 +3x-10$ are $-5$ and $2$ which should also be the zeroes of $x^3+ax^2 -7x - b$ . sub in $x=-5$ and $x=2$ we get the equations $4a - b = 6$ and $25a - b =90$ . Subtracting the first equation from the second gives us $21a = 84$ or $a=4$ . Accordingly, $4(4) - b = 6$ or $16 - b=6$ or $b =10$ . Hence, $a+b = 4+10 = \boxed{14}$ .

Let f (x)=(x-c)(x^2 +3x -10) By equating coefficients of 1) x, c=-1 2) x^2, a=4 3)constant , b=10. Hence a+b=14

Let $r_1, r_2, r_3$ be roots of $f(x)=0$ . By Vieta's Formula we have that $r_1+r_2+r_3=-a$ … (1)

$r_1 r_2 r_3=b$ … (2)

$r_1r_2+r_1r_3+r_2r_3=-7$ … (3)

We have that $f(x)$ is divisible by $x^2+3x-10$ , so, there exist $q(x)$ of degree 1, such that $(x^2+3x-10)(x-r_3)=f(x)$ . Therefore, $r_1$ and $r_2$ are roots of $x^2+3x-10=0$ and roots of $f(x)=0$ .

Then

$r_1+r_2=-3$ … (4)

$r_1r_2=-10$ … (5)

Substituting (5) at (3) we have

$-10+r_3r_1+r_3r_2=-7$

$\Longrightarrow$ $r_3(r_1+r_2)=3$

$\Longrightarrow$ $r_3=-1$

$\Longrightarrow$ $r_1+r_2+r_3=-3-1=-4=-a$

$\Longrightarrow$ $a=4$

And

$r_1r_2r_3=(-10)(-1)=10=b$

$\Longrightarrow$ $a+b=4+10=14$

$\Longrightarrow$ $\boxed{a+b=14}$

Let $P(x)=x^3+ax^2 - 7x -b$ . If $P(x)$ is divisible por $x^2 + 3x -10$ ,then the roots of $x^2+3x-10$ are also of $P(x)$

The roots of $x^2 + 3x -10=0$ are $2$ and $-5$ so the roots of $x^3+ax^2-7x-b=0$ are $2$ , $-5$ and $r$ .

By Vieta's Formula we know that

$-a = 2 +(-5) + r\\ -a=r-3\\ a=3-r$

and

$-7=2\cdot(-5) + 2r + (-5)r\\ -7=-10-3r\\ 3=-3r\\ r=-1$

Therefore $a=3-r=4$ and $b=2\cdot (-5) \cdot r = 10$

So $a+b = 14$