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for the polynomial ( p(x) = x^5 - 3x^4 - 9x + 2 ) and ( c = -1 ), find …

Question

for the polynomial ( p(x) = x^5 - 3x^4 - 9x + 2 ) and ( c = -1 ), find ( p(x) ) by (a) direct substitution and (b) the remainder theorem.

a. find ( p(-1) ) by direct substitution.
( p(-1) = square ) (type an integer.)

b. find ( p(-1) ) using the remainder theorem.
( p(-1) = square ) (type an integer.)

Explanation:

Part (a)

Step1: Substitute \( x = -1 \) into \( P(x) \)

Given \( P(x) = x^5 - 3x^4 - 9x + 2 \), substitute \( x = -1 \):

$$ LATEXBLOCK0 $$

Step2: Simplify the expression

Calculate each term:

$$ LATEXBLOCK1 $$

The Remainder Theorem states that if a polynomial \( P(x) \) is divided by \( x - c \), the remainder is \( P(c) \). Here, we want to find \( P(-1) \), which is the remainder when \( P(x) \) is divided by \( x - (-1) = x + 1 \). By the Remainder Theorem, this remainder is just \( P(-1) \), which we can calculate (or we already calculated in part (a) as 7). Alternatively, we can re - apply the substitution logic from the Remainder Theorem perspective: since the remainder when \( P(x) \) is divided by \( x + 1 \) is \( P(-1) \), and we can compute \( P(-1) \) as we did in part (a) (substituting \( x=-1\) into \( P(x) \)) to get 7.

Answer:

\( 7 \)

Part (b)