Divide using synthetic division (2x^3 + 14x^2 - 58x) / (x + 10) 58

-10 | 2 14 -58
-20 -60
2 -6 -118

So it's 2x^2 -6x The remainer is -118

Set x + 10 = 0

So that's how I get -10

I then multiplied -10 x 2 to get -20 and placed it below 14. -20 + 14 is -6. -6 x -10 is -60. -58 + -60 is -118. 2 remains the same since you didn't do anything with.

Since you started with 2x^3 and this is division, you immediately know the first term will start with x^2 so that's how I get 2x^2. The ending number, in this case -118, is always the remainder.

Step-by-step explanation:

No , theres not enough information provided to Prove that both the triangles are congruent. Here in the figure we can see that there are two triangles ∆ BDA and ∆ BDC. And its given that

AB = BC BD = BD ( common side )

The congruence conditions for two ∆s are :-

1) SAS ( Side Angle Side )

→ Two triangles are said to be congruent by SAS if two respective sides of the two triangles and the included angle between two sides are equal.

2) AAS ( Angle Angle Side )

→ Two triangles are said to be congruent by AAS if two angles and one side of triangle is congruent to other two angles and one side of the triangle .

3) SSS ( Side Side Side )

→ Two triangles are said to be congruent by SAS if all the three sides of one triangle is equal to three sides of the other triangle.

4) RHS ( Right Hypotenuse Side )

→ In two right-angled triangles, if the length of the hypotenuse and one side of one triangle, is equal to the length of the hypotenuse and corresponding side of the other triangle, then the two triangles are congruent.

And the given data doesn't satisfies any of the conditions.

Hence there is not enough information provided to Prove that two triangles are congruent .