Question - Solving for Constants in a Polynomial Function
Solution:
The image contains a question that asks you to find the values of constants 'a' and 'b' in a polynomial function $$ P(x) = 3x^3 + ax^2 + bx - 6 $$, given that $$ x - 2 $$ is a factor of $$ P(x) $$ and that the remainder is 27 when $$ P(x) $$ is divided by $$ (x + 2) $$.To solve this:1. If $$ x - 2 $$ is a factor of $$ P(x) $$, it means that $$ P(2) = 0 $$.2. If the remainder is 27 when $$ P(x) $$ is divided by $$ x + 2 $$, it means that $$ P(-2) = 27 $$.Let's use these conditions to form two equations to solve for 'a' and 'b'.1. $$ P(2) = 0 $$\[ 3(2)^3 + a(2)^2 + b(2) - 6 = 0 \]\[ 3(8) + 4a + 2b - 6 = 0 \]\[ 24 + 4a + 2b = 6 \]\[ 4a + 2b = 6 - 24 \]\[ 4a + 2b = -18 \]\[ 2a + b = -9 \] (Equation 1)2. $$ P(-2) = 27 $$\[ 3(-2)^3 + a(-2)^2 + b(-2) - 6 = 27 \]\[ 3(-8) + 4a - 2b - 6 = 27 \]\[ -24 + 4a - 2b = 27 + 6 \]\[ 4a - 2b = 33 + 24 \]\[ 4a - 2b = 57 \]\[ 2a - b = 57/2 \] (Equation 2)Now we have a system of linear equations:\[ 2a + b = -9 \] (Equation 1)\[ 2a - b = 57/2 \] (Equation 2)Multiply Equation 1 by 2 and add it to Equation 2:\[ 4a + 2b = -18 \] (Equation 1 transformed)\[ 2a - b = 57/2 \] (Equation 2)\[ (4a + 2b) + (2a - b) = -18 + 57/2 \]\[ 6a + b = 21/2 \] (Equation 3)Now subtract Equation 1 from Equation 3:\[ 6a + b - (2a + b) = 21/2 + 9 \]\[ 4a = 21/2 + 18/2 \]\[ 4a = 39/2 \]\[ a = \frac{39/2}{4} \]\[ a = 39/8 \]\[ a = 4.875 \]Now we can substitute the value of 'a' back into Equation 1 to solve for 'b':\[ 2a + b = -9 \]\[ 2(4.875) + b = -9 \]\[ 9.75 + b = -9 \]\[ b = -9 - 9.75 \]\[ b = -18.75 \]The values are $$ a \approx 4.875 $$ (or $$ a = 39/8 $$) and $$ b \approx -18.75 $$ (or $$ b = -75/4 $$).
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