We have a trinomial which is 4k, a cube minus 4k square plus 9k, and we have to take everything into account. We see that there is a common factor. We have some time if we remove that. 4k square minus 4k plus 9 and now we find the 0 of the polynomial. We apply a formula to solve 4k square minus 4k plus 9 equals 0, and we get k as 4 plus minus root of 16 minus 144 and this is divided by 8 points. This is equivalent to four plus two. We can write this in such a way that the root 128 is divided by 8 points because we know that the root of minus is 1 I. Let`s take 2 as a common factor and take 4 because 128 points means we can write this as a square 4 with 8. We get 128 points if we divide 16 by 8. The value of k is one plus one. I divide the root 8 by 2 to get the value of k which is minus 1 and minus 2. I added 2k minus 1 and 2 root 2 to get the sum.

Instead of this 4k minus 4, k plus 9 point square, it`s me. We get a result of k times 2 k minus 1 minus 2 root 2 if we take this into account completely. I added 2k minus 1 and 2 root 2 to get the sum. The answer is me and that. If possible, consider it completely. Check your answer.$2 k^{2}-22 k+$48 Factor each trinomial completely. $$ 4 k^{3}-4 k^{2}+9 k $$ Step 3: Rewrite the polynomial and divide the mean term with the two factors from step 2 above, 2 and 2 k2 + 2k + 2k + 4Step 4: Add the first 2 terms and extract similar factors: k • (k + 2) Add the last 2 terms and remove the common factors: 2 • (k + 2) Step 5: Add the four terms of Step 4: (k+2) • (k+2) What is the desired factorization 3.3 Find roots (zeros) of: F(k) = k3+10k2+28k+24 See theory in step 3.1 In this case, the principal coefficient is 1 and the end constant is 24. The factors are: the main coefficient: 1 of the roulette constant: 1, 2, 3, 4, 6, 8, 12, 24 Testons. Find the polynomial that counts as $(4 k+9)(k+2)$. 3.6 Multiply (k+2) by (k+2) The rule says: To multiply exponential expressions that have the same basis, add their exponents.

In our case, the common basis is (k+2) and the exponents are: 1 , since (k+2) is the same number as (k+2)1 and 1 since (k+2) is the same number as (k+2)1. The product is therefore (k+2)(1+1) = (k+2)2. 3.1 Find roots (zeros) of: F(k) = k4+8k3+8k2-32k-48The polynomial root calculator is a set of methods for finding values of k for which F(k)=0 Rational Roots Test is one of the above tools. He would find only rational roots, which are numbers k, which can be expressed as the quotient of two integers, the rational root theorem states that if a polynomial is zero for a rational number P/Q, then P is a factor of the end constant and Q is a factor of the principal coefficient, In this case, the principal coefficient is 1 and the end constant is -48. The factors are: the main coefficient: 1 of the roulette constant: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 Testons. The factor theorem states that if P/Q is the root of a polynomial, that polynomial can be divided by q * x-p Note that q and p come from P/Q, reduced to its lowest terms. In our case, this means that k4 + 8k3 + 8k2-32k-48 can be divided by 3 different polynomials, including by k-2 3.5 factorization k2 + 4k + 4 The first term is, K2 Its coefficient is 1. The medium term is, +4k its coefficient is 4. The last term, « the constant », is +4 Step 1: Multiply the coefficient of the first term by the constant 1 • 4 = 4 Step 2: Find two factors of 4 whose sum is equal to the coefficient of the mean term, which is 4. 3.2 Long polynomial division: k4+8k3+8k2-32k-48 (« dividend ») by: k-2 (« divisor »).