This will be a bit hard to write out here, but let's see. I'm assuming that this is a typo. You may have meant 3x3. If it was 3x2, you wouldn't need synthetic division and you could just factor this to get x=1,x=−5.
Assuming 3x3... First, you let the coefficients of each degree be used in the division (3,−1,8,−10).
Then, dividing by x−1 implies that you use 1 in your upper left (if it was x+1, put −1). So, draw the bottom and right sides of a square, put 1 inside it, and then write 3 -1 8 -10 to the right.
1∣∣ 3 -1 8 -10
+
−−−−−
First, bring the 3 down to the bottom, and multiply it by the 1. Put that 3 below −1.
1∣∣ 3 -1 8 -10
+ 3
−−−−−
3
Then add it up:
1∣∣ 3 -1 8 -10
+ 3
−−−−−
3 2
Repeat a few times once you've figured out the simple pattern (divisor⋅new sum, put result under next-lowest degree term, add to get another new sum, repeat).
1∣∣ 3 -1 8 -10
+ 3 2
−−−−−
3 2 10
(All that really happened here was 1⋅2=2 and 8+2=10.)
1∣∣ 3 -1 8 -10
+ 3 2 10
−−−−−
3 2 10 0
(Then to finish it up, 1⋅10=10, and −10+10=0.)
You know you can stop when you reach the far right and you have no spot left below the original dividend (3 -1 8 -10) to insert a product.
Since presumably you started with a cubic, the answer is a quadratic. Thus, you have:
3x2+2x+10x0+(r=0)
Indeed, if you multiply them together, you get the original back:
(3x2+2x+10)(x−1)=3x3−3x2+2x2−2x+10x−10=3x3−x2+8x−10
So the answer would be 3x2+2x+10. In a general case, you could write it like this:
Let f(x)=ax3+bx2+cx+d and g(x)=x−h. If the solution is h(x), and a remainder is r(x), then:
h(x)=f(x)g(x)+r(x)g(x)
So you would have:
h(x)=answer=3x3−x2+8x−10x−1=3x2+2x+10+r(x)x−1
where r(x)=0.
Had you meant 3x2−x2+8x−10, you could have just done the quadratic equation. Or, factor.
3x2−x2+8x−10=2x2+8x−10⇒(2x−2)(x+5)=0
⇒x=1,x=−5
