+23246 The polynomial x3 + ax2 - bx = 6
---> x3 + ax2 - bx - 6 = 0
Since it has the factors (x + 1)(x + 2), we need to find the third factor.
(x + 1)(x + 2) ---> x2 + 3x + 2
The third factor must have just x for the x-term (to get a final product of x3)
and a number factor of -3 (to get a final product of -6).
The third factor is: x - 3.
Multiplying the factors (x + 1)(x + 2)(x - 3), we get: x3 + 0x2 - 7x - 6.
Therefore, the value of a is 0 and the value of b is -7.
+26367 the polynomial x^3+ax^2-bx=6 has factors of (x+1) (x+2) what is a and b
I. factors of \((x+1) \quad \Rightarrow x_1 = -1\)
\(\begin{array}{|rcll|} \hline x^3+ax^2-bx -6 &=& 0 \quad &| \quad x_1 = -1\\ ( -1)^3+a( -1)^2-b(-1) -6 &=& 0 \\ -1+a+b -6 & =& 0 \\ a+b -7 & =& 0 \quad &| \quad +7\\ a+b & =& 7 \\ \mathbf{ b} & \mathbf{=}& \mathbf{7-a} \\ \hline \end{array}\)
II. factors of (x+2) \((x+2) \quad \Rightarrow x_2 = -2\)
\(\begin{array}{|rcll|} \hline x^3+ax^2-bx -6&=& 0 \quad &| \quad x_2 = -2\\ ( -2)^3+a( -2)^2-b(-2) -6 &=& 0 \\ -8+4a+2b -6 &=& 0 \\ 4a+2b -14 &=& 0 \quad &| \quad +14\\ 4a+2b &=& 14 \quad &| \quad : 2\\ 2a+ b &=& 7 \quad &| \quad \mathbf{ b = 7-a} \\ 2a+ 7-a &=& 7 \\ a+ 7 &=& 7 \quad &| \quad x_2 = -7\\ a &=&0 \\ \mathbf{ a} & \mathbf{=}& \mathbf{0} \\\\ b &=& 7-a \quad &| \quad \mathbf{ a} \mathbf{=} \mathbf{0}\\ b &=& 7 - 0 \\ b &=& 7\\ \mathbf{ b} & \mathbf{=}& \mathbf{7 } \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline x^3+ax^2-bx &=& 6 \quad &| \quad \mathbf{ a} \mathbf{=} \mathbf{0} \qquad \mathbf{ b = 7 }\\ x^3+0\cdot x^2- 7\cdot x &=& 6 \\ \mathbf{x^3 - 7\cdot x }& \mathbf{=}& \mathbf{6} \\ \hline \end{array}\)
