Write the polynomial as the product of linear factors and list all the zeros of the function. g(x)=x3−3x2+x+5

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Jeffery Autrey · Accepted Answer

Step 1  Given: The polynomial function g(x)=x3−3x2+x+5,  To determine: The factorization and zeros of polynomial  Step 2  Explanation:  Factorizing the given polynomial function:  Here it can be observed that   g(−1)=(−1)3−3(−1)2+(−1)+5  ⇒g(−1)=−1−3−1+5=0  Therefore (x+1) is a factor of the polynomial x3−3x2+x+5,  Therefore polynomial x3−3x2+x+5 can be written as  x3−3x2+x+5=x2(x+1)−4x(x+1)+5(x+1)  ⇒x3−3x2+x+5=(x+1)(x2−4x+5)  Here x2−4x+5 is irreducible.  Therefore given polynomial function can be factorized as   g(x)=(x+1)(x2−4x+5)  Step 3  Finding the zeros of the given polynomial function:  Solving g(x)=0 for x,  g(x)=0  ⇒(x+1)(x2−4x+5)=0  ⇒x=−1    [here x2−4x+5≠0]  Therefore x=-1 is the only zero of the given polynomial function.

Nick Camelot · Accepted Answer

We can start by using the Rational Root Theorem to identify potential rational zeros. According to the theorem, the possible rational zeros of g(x) are of the form ±pq, where p is a factor of the constant term (in this case, 5) and q is a factor of the leading coefficient (in this case, 1).The factors of 5 are ±1 and ±5, while the factors of 1 are ±1. Therefore, the possible rational zeros of g(x) are ±1 and ±5.Next, we can use synthetic division or long division to check these potential zeros. Let&#039;s start with x=1:11−3151−2−1Since the remainder is not zero, x=1 is not a zero of g(x). Let&#039;s try x=−1:−11−315−14−5Again, the remainder is not zero, so x=−1 is not a zero either. Let&#039;s continue with x=5:51−31551055Once again, the remainder is not zero, meaning x=5 is not a zero. Finally, let&#039;s check x=−5:−51−315−540−195The remainder is zero, which indicates that x=−5 is a zero of g(x).Now, we can factor g(x) using the zero we found:g(x)=(x−(−5))(x2+5x+39)The quadratic factor x2+5x+39 does not have real roots, so we cannot further factor it using linear factors. Therefore, the polynomial g(x) expressed as a product of linear factors is:g(x)=(x+5)(x2+5x+39)The zeros of g(x) are x=−5 (with multiplicity 1) and the complex zeros of the quadratic factor x2+5x+39.

Eliza Beth13 · Accepted Answer

Step 1:To write the polynomial as the product of linear factors and list all the zeros of the function g(x)=x3−3x2+x+5, we can use the factor theorem and synthetic division.First, we find the possible rational zeros by considering the factors of the constant term (5) divided by the factors of the leading coefficient (1). The possible rational zeros are ±1 and ±5.Using synthetic division, we test these possible zeros. By substituting each value into the polynomial, we find that x=−1 is a zero. Performing synthetic division with x=−1, we get:−11−315−14−5Step 2:The result shows that x=−1 is a zero of g(x), and the resulting quotient is x2−x+5. We can now write the polynomial as the product of linear factors:g(x)=(x+1)(x2−x+5)To find the remaining zeros, we solve the quadratic factor x2−x+5 by factoring or using the quadratic formula. However, in this case, the quadratic factor has no real roots since the discriminant (b2−4ac) is negative.Therefore, the zeros of g(x) are x=−1 and two complex roots from the quadratic factor.

Write the polynomial as the product of linear factors and

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