dividend x3 - 11x2 + 16x + 84
- divisor * x2 x3 - 7x2
remainder - 4x2 + 16x + 84
- divisor * -4x1 - 4x2 + 28x
remainder - 12x + 84
- divisor * -12x0 - 12x + 84
remainder 0
Quotient : x2-4x-12 Remainder: 0
Step-1 : Multiply the coefficient of the first term by the constant 1 • -12 = -12
Step-2 : Find two factors of -12 whose sum equals the coefficient of the middle term, which is -4 .
-12 + 1 = -11
-6 + 2 = -4 That's it
Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above, -6 and 2
x2 - 6x + 2x - 12
Step-4 : Add up the first 2 terms, pulling out like factors :
x • (x-6)
Add up the last 2 terms, pulling out common factors :
2 • (x-6)
Step-5 : Add up the four terms of step 4 :
(x+2) • (x-6)
Which is the desired factorization
Final result :
(x + 2) • (x - 6) • (x - 7)
answer from Kalahira
x + 4Further explanationGiven:The expression x³ - 11x² + 16x + 84Question:Which binomial is not a divisor of x³ - 11x² + 16x + 84?x + 2x + 4x - 6x - 7The Process:This is a problem about Remainder Theorem. If x = a or (x - a) is one of the roots or factors (divisors) of the function f (x), then f(a) = 0.Let us test one by one of the options available into the polynomial function.[tex]\boxed{ \ x + 2 \ } \rightarrow \boxed{ \ x = -2 \ }.[/tex][tex]\boxed{ \ \rightarrow (-2)^3 - 11(-2)^2 + 16(-2) + 84 = ? \ }[/tex][tex]\boxed{ \ \rightarrow -8 - 44 - 32 + 84 = 0 \ }[/tex] [tex]\boxed{ \ x + 4 \ } \rightarrow \boxed{ \ x = -4 \ }.[/tex][tex]\boxed{ \ \rightarrow (-4)^3 - 11(-4)^2 + 16(-4) + 84 = ? \ }[/tex][tex]\boxed{ \ \rightarrow -64 - 176 - 32 + 84 = -188 \ }[/tex][tex]\boxed{ \ x - 6 \ } \rightarrow \boxed{ \ x = 6 \ }.[/tex][tex]\boxed{ \ \rightarrow (6)^3 - 11(6)^2 + 16(6) + 84 = ? \ }[/tex][tex]\boxed{ \ \rightarrow 216 - 396 + 96 + 84 = 0 \ }[/tex][tex]\boxed{ \ x - 7 \ } \rightarrow \boxed{ \ x = 7 \ }.[/tex][tex]\boxed{ \ \rightarrow (7)^3 - 11(7)^2 + 16(7) + 84 = ? \ }[/tex][tex]\boxed{ \ \rightarrow 343 - 539 + 112 + 84 = 0 \ }[/tex]From the test results above, it was clearly observed that (x + 4) is not a divisor of x³ - 11x² + 16x + 84 because [tex]\boxed{ \ f (-4) \neq 0 \ }[/tex].- - - - - - - - - -Alternative StepsWe will use Horner's rule for polynomial division in finding factors as well as divisors.To begin, let us try x = -2 to see if it is one of the roots of x³ - 11x² + 16x + 84. The complete process can be seen in the attached picture.Since there is no remainder when x³ - 11x² + 16x + 84 is divided by (x + 2), this binomial is a divisor (or a factor). The quotient is obtained x² - 13x + 42. After factorization, we get (x - 6) dan (x - 7 as factors and also the divisor.Learn moreThe remainder theorem https://brainly.com/question/9500387A polynomial of the 5th degree with a leading coefficient of 7 and a constant term of 6 https://brainly.com/question/12700460 The product of a binomial and a trinomial is x 3 + 3 x 2 − x + 2 x 2 + 6 x − 2.
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