# Show that $-1$ is a root of the equation.

Show that $-1$ is a root of the equation $$(a+b-2c)x^2+(2a-b-c)x+(c+a-2b)=0$$ Find the other root.

My Attempt:

Since replacing $x=-1$ satisfies the given equation, it is the root. But how to find the other root.

• Hint: product of the roots is... – dxiv Aug 17 '17 at 23:24
• Vieta's relations. – Bernard Aug 17 '17 at 23:25
• ... and the sum of the roots is ... – Henry Aug 17 '17 at 23:25
• Given a quadratic equation $px^2+qx+r=0$, the sum of the roots is $-q/p$ and the product is $r/p$. – Oscar Lanzi Aug 17 '17 at 23:25
• @dxiv, Sum of the roots is $\dfrac {b+c-2a}{a+b-2c}$. – pi-π Aug 17 '17 at 23:26

A few people in the comments are citing Vieta's rlations, and that's a good strategy. There's also this: If $-1$ is a root, then $(x+1)$ is a factor. Use polynomial long division to divide the given polynomial by $(x+1)$, set the linear quotient equal to zero, and solve.
Suppose you have something like $8x^2 + 17x+9,$ and you plug in a number, such as $-1,$ and you find that $8(-1)^2 + 17(-1) + 9 =0.$ If you plug a number into a polynomial and get $0$, that means $x$ minus that number is a factor, so in this case you have $x+1$ is a factor: $$8x^2 + 17x+9 = (x+1)(\cdots\cdots\cdots).$$ If all else fails, you can find the other factor by long division: In this case, divide $8x^2+17x+9$ by $x+1,$ getting $8x+9.$