Determine whether the given binomial is a factor of the polynomial \( p(x) \). If it is, find the remaining factors of \( p(x) \). 2. \( p(x)=x^{3}-2 x^{2}+7 x-14 ; x-2 \)
Real Tutor Solution
Answer
Solution
Reviewed and approved by the UpStudy tutoring team
Bonus Knowledge
To determine if \( x - 2 \) is a factor of the polynomial \( p(x) = x^{3} - 2x^{2} + 7x - 14 \), you can use the Factor Theorem, which asserts that if \( p(c) = 0 \) for some value \( c \), then \( x - c \) is a factor of \( p(x) \). Substituting \( x = 2 \) into \( p(x) \): \[ p(2) = (2)^{3} - 2(2)^{2} + 7(2) - 14 \] \[ = 8 - 8 + 14 - 14 = 0 \] Since \( p(2) = 0 \), \( x - 2 \) is indeed a factor. To find the remaining factors of \( p(x) \), you can perform polynomial long division of \( p(x) \) by \( x - 2 \). Performing the long division, we find: \[ p(x) \div (x - 2) = x^{2} + 0x + 7 = x^{2} + 7 \] Thus, we have \[ p(x) = (x - 2)(x^{2} + 7). \] The remaining factor \( x^{2} + 7 \) can’t be factored further using real numbers, but it has complex roots. Therefore, the complete factorization over the complex numbers would be: \[ p(x) = (x - 2)(x - i\sqrt{7})(x + i\sqrt{7}). \] Now, if you're feeling adventurous with polynomial factors and want to dive deeper, remember that factoring can be visualized like puzzle solving! Each polynomial can hide its secrets behind cleverly arranged terms, just waiting for a keen eye (and perhaps some division) to unravel the mystery of its composition. When working through polynomial long division, it's essential to keep your terms organized. A common mistake is to misalign degrees or skip steps, leading to confusion. Think of it as balancing a scale: every term must have its place to keep the equation's integrity intact!