<span>Consider a angle â BAC and the point D on its defector
Assume that DB is perpendicular to AB and DC is perpendicular to AC.
Lets prove DB and DC are congruent (that is point D is equidistant from sides of an angle â BAC
Proof
Consider triangles ΔADB and ΔADC
Both are right angle, â ABD= â ACD=90 degree
They have congruent acute angle â BAD and â CAD( since AD is angle bisector)
They share hypotenuse AD
therefore these right angle are congruent by two angle and sides and, therefore, their sides DB and DC are congruent too, as luing across congruent angles</span>
Given:

To find:
The product of the polynomials.
Solution:
1.

Multiply the numerical coefficient and add the powers of x.

2. 
Multiply each term of first polynomial with each term of 2nd polynomial.
Multiply the numerical coefficient and add the powers of x.


3. 
Multiply each term of first polynomial with each term of 2nd polynomial.
Multiply the numerical coefficient and add the powers of x.

Add or subtract like terms together.

The answer for multiplying polynomials:



Step-by-step explanation:
The dividend is the number which is being divided and the divisor is what you divided by. An example is:
12 ÷ 3. The quotient is 4.
Answer:32
Step-by-step explanation:firs we multiply 5*4 =20 .then we subtract 2-1=1 add those numbers 20+1=21 and then add 11 which equals 32