Construct an isosceles right-angled triangle ∆PQR where ∠

Please find attached the required triangle drawn with Microsoft Visio

Step-by-step explanation:

The isosceles triangle ΔPQR we are to construct has the following parameters;

∠PQR = 90°

Length of side hypotenuse PR = 4 cm

Therefore, given that the triangle is right triangle is an isosceles triangle the length of each of the side legs are equal

Therefore, we have;

Length of PQ = Length of QR

The length of the legs , and , are given by Pythagoras's theorem as follows;

= √(² + ²) = √(² + ²) = ·√2

∴ = 4 cm/(√2) = 2·√2 cm =

The required triangle drawn with Microsoft Visio is attached

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0