hello, please help me solve to find the correct polynomials!

INFORMATION: We have the next polynomialsAnd we must factor them to complete the next tableSTEP BY STEP EXPLANATION: 1. [tex]x^2-8x+15[/tex]To factor it, we must look for two number that multiplied be equal to 15 and added up be equal to -8.These two numbers would be -5 and -3.- -5 x -3 = 15- -5 - 3 = -8So, when we factor this polynomial, we obtain[tex]\begin{gathered} x^2-8x+15=(x-5)(x-3) \\ \text{ So, }a=1,b=-5,c=1,d=-3 \end{gathered}[/tex]2. [tex]2x^3-8x^2-24x[/tex]To factor it, we must first take the common factor 2x from the expression[tex]2x(x^2-4x-12)[/tex]Now, we must factor the terms in the parenthesis. We must look for two number that multiplied be equal to -12 and added up be equal to -4. These two numbers would be -6 and 2.- -6 x 2 = -12- -6 + 2 = -4So, when we factor this polynomial, we obtain[tex]\begin{gathered} 2x(x+2)(x-6) \\ \text{ So, }a=1,b=2,c=1,d=-6 \end{gathered}[/tex]3. [tex]6x^2+14x+4[/tex]To factor it, we must first take the common factor 2 from the expression[tex]2(3x^2+7x+2)[/tex]Then, we divide the 7x term in the parenthesis in two terms[tex]2(3x^2+6x+x+2)[/tex]Now, we can take the common factor x + 2 in the parenthesis[tex]2(3x(x+2)+(x+2))[/tex]Finally, we can take the common factor x + 2 in the complete expression[tex]\begin{gathered} 2(x+2)(3x+1) \\ \text{ Simplifying,} \\ =\left(3x+1\right)(2x+4) \\ \text{ So, }a=3,b=1,c=2,d=4 \end{gathered}[/tex]ANSWER: