Prove that when a polynomial function p(x) is divided by a first degree polynomial ax + b the remainder is p(-b/a)

r = p(-b/a)

Step-by-step explanation:

Let p(x) = q(x)(ax + b) + r(x) where q(x) is the quotient when p(x) is divided by ax + b and r(x) is the remainder.

Since ax + b is a first degree polynomial, r(x) is one power less than ax + b is is just a constant, r.

So, p(x) = q(x)(ax + b) + r

Now, p(x) = r when q(x)(ax + b) = 0

since q(x) ≠ 0, ax + b = 0 ⇒ ax = -b ⇒ x = -b/a

⇒ p(x) = r when x = -b/a

So, r = p(-b/a)

So, the remainder when a polynomial function p(x) is divided by a first degree polynomial ax + b is p(-b/a)

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

step-by-step explanation:

given   that triangle lno is congruent to triangle lnm.

we have to fill up the blanks in the given question.

according to above order of congruency   kept in mind we can say that

∠lno ≅ ∠lnm

∠oln ≅ ∠mln

ln ≅ ln due to common in both.

hence, by sas congruency,

the side angle side rule   states that if two sides and the included angle of one triangle are congruent to another triangle, then these two triangles are congruent.