Answer:
1) The factorized form of the polynomial is [tex](x-5)\cdot (x+2) = 0[/tex].
2) The factorized form of the polynomial is [tex](x+6)\cdot (x-4) = 0[/tex].
3) [tex]r_{1} = -3[/tex], [tex]r_{2} = -2[/tex]. (Option D)
4) [tex]r_{1} = -7[/tex], [tex]r_{2} = 5[/tex]. (Option A)
Step-by-step explanation:
All exercise are case of factorization of second grade polynomials of the form [tex]x^{2} +(-r_{1} - r_{2})\cdot x + r_{1}\cdot r_{2}[/tex], where [tex]r_{1}[/tex] and [tex]r_{2}[/tex] are the two roots of the polynomial. Now we proceed to solve each polynomial:
1) [tex]x^{2}-3\cdot x-10 = 0[/tex]
In this case, the coefficients have the following characteristics:
[tex]-r_{1}-r_{2} = -3[/tex]
[tex]r_{1}\cdot r_{2} = -10[/tex]
The solution of this system of nonlinear equations is: [tex]r_{1} = 5[/tex], [tex]r_{2} = -2[/tex].
Then, the factorized form of the polynomial is:
[tex](x-5)\cdot (x+2) = 0[/tex]
2) [tex]x^{2}+2\cdot x -24 = 0[/tex]
In this case, the coefficients have the following characteristics:
[tex]-r_{1}-r_{2} = 2[/tex]
[tex]r_{1}\cdot r_{2} = -24[/tex]
The solution of this system of nonlinear equations is: [tex]r_{1} = -6[/tex], [tex]r_{2} = 4[/tex].
Then, the factorized form of the polynomial is:
[tex](x+6)\cdot (x-4) = 0[/tex]
3) [tex]x^{2} + 5\cdot x +6 = 0[/tex]
In this case, the coefficients have the following characteristics:
[tex]-r_{1}-r_{2} = 5[/tex]
[tex]r_{1}\cdot r_{2} = 6[/tex]
The solution of this system of nonlinear equations is: [tex]r_{1} = -3[/tex], [tex]r_{2} = -2[/tex].
Hence, the correct answer is D.
4) [tex]x^{2}+2\cdot x -35 = 0[/tex]
In this case, the coefficients have the following characteristics:
[tex]-r_{1}-r_{2} = 2[/tex]
[tex]r_{1}\cdot r_{2} = -35[/tex]
The solution of this system of nonlinear equations is: [tex]r_{1} = -7[/tex], [tex]r_{2} = 5[/tex].
Hence, the correct answer is A.
