3. If \( 3 x^{3}-a x^{2}+2 x+1 \) has a remainder of -3 on division by \( x+2 \), find a. 4. Find both values of ' \( a \) ' for which the polynomials \( 3 x^{2}-7 x-4 \) and \( 32 x^{2}-10 x+6 \) give the same remainder on division by \( x-a \).

Solve the equation by following steps: - step0: Solve for \(a\): \(3\left(-2\right)^{3}-a\left(-2\right)^{2}+2\left(-2\right)+1=-3\) - step1: Simplify: \(-27-4a=-3\) - step2: Move the constant to the right side: \(-4a=-3+27\) - step3: Add the numbers: \(-4a=24\) - step4: Change the signs: \(4a=-24\) - step5: Divide both sides: \(\frac{4a}{4}=\frac{-24}{4}\) - step6: Divide the numbers: \(a=-6\) Solve the system of equations \( 3x^{2}-7x-4=R; 32x^{2}-10x+6=R \). Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}3x^{2}-7x-4=R\\32x^{2}-10x+6=R\end{array}\right.\) - step1: Rewrite the expression: \(\left\{ \begin{array}{l}R=3x^{2}-7x-4\\32x^{2}-10x+6=R\end{array}\right.\) - step2: Substitute the value of \(R:\) \(32x^{2}-10x+6=3x^{2}-7x-4\) - step3: Move the expression to the left side: \(32x^{2}-10x+6-\left(3x^{2}-7x-4\right)=0\) - step4: Calculate: \(29x^{2}-3x+10=0\) - step5: Solve using the quadratic formula: \(x=\frac{3\pm \sqrt{\left(-3\right)^{2}-4\times 29\times 10}}{2\times 29}\) - step6: Simplify the expression: \(x=\frac{3\pm \sqrt{\left(-3\right)^{2}-4\times 29\times 10}}{58}\) - step7: Simplify the expression: \(x=\frac{3\pm \sqrt{-1151}}{58}\) - step8: The expression is undefined: \(x \notin \mathbb{R}\) - step9: The system of equations has no solution in the set of real numbers: \(\left(R,x\right) \notin \mathbb{R}^{2}\) - step10: Alternative Form: \(\textrm{No real solution}\) Let's solve the problems step by step. ### Problem 3 We need to find the value of \( a \) such that the polynomial \( 3x^{3} - ax^{2} + 2x + 1 \) has a remainder of -3 when divided by \( x + 2 \). According to the Remainder Theorem, the remainder of a polynomial \( f(x) \) when divided by \( x - c \) is \( f(c) \). Here, we need to evaluate the polynomial at \( x = -2 \): \[ f(-2) = 3(-2)^{3} - a(-2)^{2} + 2(-2) + 1 \] Calculating \( f(-2) \): \[ f(-2) = 3(-8) - a(4) - 4 + 1 = -24 - 4a - 4 + 1 = -27 - 4a \] Setting this equal to -3 (the given remainder): \[ -27 - 4a = -3 \] Now, solving for \( a \): \[ -4a = -3 + 27 \] \[ -4a = 24 \] \[ a = -6 \] Thus, the value of \( a \) is \( -6 \). ### Problem 4 We need to find the values of \( a \) for which the polynomials \( 3x^{2} - 7x - 4 \) and \( 32x^{2} - 10x + 6 \) give the same remainder when divided by \( x - a \). Let the remainder when dividing by \( x - a \) be \( R \). We can express this as: 1. \( 3a^{2} - 7a - 4 = R \) 2. \( 32a^{2} - 10a + 6 = R \) Setting these two equations equal to each other: \[ 3a^{2} - 7a - 4 = 32a^{2} - 10a + 6 \] Rearranging gives: \[ 3a^{2} - 32a^{2} + 10a - 7a - 4 - 6 = 0 \] \[ -29a^{2} + 3a - 10 = 0 \] This is a quadratic equation, but upon solving, we find that there are no real solutions. Therefore, there are no values of \( a \) for which the two polynomials give the same remainder when divided by \( x - a \). ### Summary of Results - For Problem 3, \( a = -6 \). - For Problem 4, there are no real solutions for \( a \).