The polynomial p(x)=x^(3)-19 x-30 has a known factor of (x+2). Rewrite p(x) as a product of linear factors. p(x)=

Factor Finding: Since we know that (x + 2) is a factor of p(x), we can perform polynomial long division or synthetic division to divide p(x) by (x + 2) to find the other factors.Synthetic Division: Let's use synthetic division to divide p(x) by (x + 2). We set up the synthetic division with -2 (the zero of the factor x + 2) and the coefficients of p(x): 1 (for x^3), 0 (for x^2, since it is missing), -19 (for x), and -30 (for the constant term). -2 | 1 0 -19 -30 | -2 4 30 ----------------- 1 -2 -15 0 The remainder is 0, which confirms that (x + 2) is indeed a factor. The other coefficients give us the quotient polynomial: x^2 - 2x - 15.Quadratic Polynomial: Now we need to factor the quadratic polynomial x^2 - 2x - 15. We look for two numbers that multiply to -15 and add to -2. These numbers are -5 and 3.Factoring: We can now write x^2 - 2x - 15 as (x - 5)(x + 3). So, the polynomial p(x) can be written as the product of its linear factors: p(x) = (x + 2)(x - 5)(x + 3).

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The polynomial
p(x)=x^(3)-19 x-30 has a known factor of
(x+2).
Rewrite
p(x) as a product of linear factors.

p(x)=

The polynomial $p(x)=x^{3}-19 x-30$ has a known factor of $(x+2)$ . $\newline$ Rewrite $p(x)$ as a product of linear factors. $\newline$ $p(x)=$

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Math Problems

Algebra 1

Find the inverse of a linear function

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Q. The polynomial

p(x)=x^{3}-19 x-30

has a known factor of

(x+2)

\newline

Rewrite

p(x)

as a product of linear factors.

\newline

p(x)=

Factor Finding: Since we know that $(x + 2)$ is a factor of $p(x)$ , we can perform polynomial long division or synthetic division to divide $p(x)$ by $(x + 2)$ to find the other factors.
Synthetic Division: Let's use synthetic division to divide $p(x)$ by $(x + 2)$ . We set up the synthetic division with $-2$ (the zero of the factor $x + 2$ ) and the coefficients of $p(x)$ : $1$ (for $x^3$ ), $0$ (for $x^2$ , since it is missing), $-19$ (for $(x + 2)$ $0$ ), and $(x + 2)$ $1$ (for the constant term). $\newline$ $-2$ | $1$ $0$ $-19$ $(x + 2)$ $1$ $\newline$ | $-2$ $(x + 2)$ $8$ $(x + 2)$ $9$ $\newline$ ----------------- $\newline$ $1$ $-2$ $-2$ $2$ $0$ $\newline$ The remainder is $0$ , which confirms that $(x + 2)$ is indeed a factor. The other coefficients give us the quotient polynomial: $-2$ $6$ .
Quadratic Polynomial: Now we need to factor the quadratic polynomial $x^2 - 2x - 15$ . We look for two numbers that multiply to $-15$ and add to $-2$ . These numbers are $-5$ and $3$ .
Factoring: We can now write $x^2 - 2x - 15$ as $(x - 5)(x + 3)$ . So, the polynomial $p(x)$ can be written as the product of its linear factors: $p(x) = (x + 2)(x - 5)(x + 3)$ .