Find real zeros of $p(x)=x^3-4x^2+6x-24$

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Algebra 2

Quadratic equation with complex roots

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Q. Find real zeros of

p(x)=x^3-4x^2+6x-24

Check for Factors: Let's try to factor by grouping or synthetic division, but first, let's check if there are any obvious factors.
Rational Root Theorem: We can use the Rational Root Theorem to list possible rational zeros, which are $\pm$ factors of $24$ (constant term) divided by factors of $1$ (leading coefficient). So, possible zeros are $\pm1$ , $\pm2$ , $\pm3$ , $\pm4$ , $\pm6$ , $\pm8$ , $\pm12$ , $24$ $0$ .
Test Possible Zeros: Let's test these possible zeros using synthetic division or direct substitution. We'll start with $1$ . $p(1) = 1^3 - 4(1)^2 + 6(1) - 24 = 1 - 4 + 6 - 24 = -21$ , which is not zero.
Synthetic Division with $4$ : Let's try $2$ . $p(2) = 2^3 - 4(2)^2 + 6(2) - 24 = 8 - 16 + 12 - 24 = -20$ , which is not zero.
Factor Out $(x - 4)$ : Let's try $3$ . $p(3) = 3^3 - 4(3)^2 + 6(3) - 24 = 27 - 36 + 18 - 24 = -15$ , which is not zero.
Perform Synthetic Division: Let's try $4$ . $\newline$ $p(4) = 4^3 - 4(4)^2 + 6(4) - 24 = 64 - 64 + 24 - 24 = 0$ , so $4$ is a zero.
Find $b$ and $c$ : Now we can factor out $(x - 4)$ from $p(x)$ using synthetic division or long division. $\newline$ $p(x) = (x - 4)(x^2 + bx + c)$ , we need to find $b$ and $c$ .
Quotient is $x^2 + 6$ : Performing synthetic division with $4$ , we get: $\newline$ $4 | 1 -4 6 -24$ $\newline$ $| 4 0 24$ $\newline$ ---------------- $\newline$ $1 0 6 0$ $\newline$ So, the quotient is $x^2 + 0x + 6$ , which simplifies to $x^2 + 6$ .
Final Polynomial: Now we have $p(x) = (x - 4)(x^2 + 6)$ . The quadratic $x^2 + 6$ doesn't have real roots because it's always positive.
Real Zero of $p(x)$ : The only real zero of $p(x)$ is $x = 4$ .