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WARMUP 2 (

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Presentation on theme: "WARMUP 2 ("β€” Presentation transcript:

1 WARMUP 2 (π‘₯ 3 +4 π‘₯ 2 βˆ’15π‘₯βˆ’18)Γ·(π‘₯βˆ’3)

2 4-8 The Remainder and Factor Theorems
Evaluate FUNCTIONS USING SYNTHETIC SUBSTITUTION Determine whether a binomial is a factor of A POLYNOMIAL BY USING SYNTHETIC SUBSTITUTION

3 Division of polynomials is useful in trying to factor polynomials
Division of polynomials is useful in trying to factor polynomials. As in our warmup, a remainder of zero tells you that the factor divided in exactly with no remainder, so you have just split your polynomial up into at least two factors. If we can reduce the polynomial to a quadratic, we can ALWAYS find the remaining factors, even if they involve complex numbers or radicals. (Quadratic formula)

4 Today we will look at using division to find factors
Today we will look at using division to find factors. It is NOT always possible to factor without knowing where to start looking. 17. Given a polynomial and one of its factors, find the remaining factors of the polynomial. π‘₯ 3 βˆ’3π‘₯+2; π‘₯+2 Was this an equation? Are there zeros?

5 Given a polynomial and one of its factors, find the remaining factors.
π‘₯ π‘₯ 2 +23π‘₯ βˆ’42; π‘₯ βˆ’1

6 Sometimes we would like to know a function value without doing direct substitution. We can use synthetic substitution to accomplish the same thing. Synthetic substitution is using the process of synthetic division to find out what the remainder is when dividing by linear factor π‘₯βˆ’π‘Ÿ. The remainder of the synthetic division is the y-value for the function when r is substituted in.

7 Which is fastest? 11. Find f(-5) and f(2) for 𝑓 π‘₯ = 2π‘₯ 3 βˆ’8 π‘₯ 2 βˆ’2π‘₯+5
Compare these two processes: Regular substitution vs synthetic substitution Which is fastest?

8 Q(x) P(x) Divisor Remainder = P(r)

9

10 Use the graph to find all of the factors for the polynomial function.

11 27.

12 Challenge: Find the solutions of the polynomial equation.


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