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# Find if the expansion of the product of and has no term.

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Find  if the expansion of the product of  and  has no  term.

Dec 4, 2014

#1
+20850
+10

Find  if the expansion of the product of  and  has no  term.

$$\small{\text{  (x^3 - 4x^2 + 2x - 5)*( x^2 + tx - 7) = tx^4-4tx^3+{2tx^2}-5tx+x^5-4x^4-5x^3+{23x^2}-14x+35  }}$$

$$\\ \small{\text{ Set  {2tx^2}+{23x^2} = 0  than the product has no x^2, t must be a constant! }} \\\\ 2tx^2 + 23x^2 = 0 \\\\ 2tx^2 = - 23x^2 \quad | \quad : 2x^2 \\\\ t= -\frac{23}{2} = - 11.5$$

$$\small{\text{  {t= -11.5}\qquad (x^3-4x^2+2x-5)*(x^2+{(-11.5)}x-7) = x^5 - 15.5x^4+41x^3+43.5x+35  }}$$

There is no more $$x^2$$

.
Dec 4, 2014

#1
+20850
+10

Find  if the expansion of the product of  and  has no  term.

$$\small{\text{  (x^3 - 4x^2 + 2x - 5)*( x^2 + tx - 7) = tx^4-4tx^3+{2tx^2}-5tx+x^5-4x^4-5x^3+{23x^2}-14x+35  }}$$

$$\\ \small{\text{ Set  {2tx^2}+{23x^2} = 0  than the product has no x^2, t must be a constant! }} \\\\ 2tx^2 + 23x^2 = 0 \\\\ 2tx^2 = - 23x^2 \quad | \quad : 2x^2 \\\\ t= -\frac{23}{2} = - 11.5$$

$$\small{\text{  {t= -11.5}\qquad (x^3-4x^2+2x-5)*(x^2+{(-11.5)}x-7) = x^5 - 15.5x^4+41x^3+43.5x+35  }}$$

There is no more $$x^2$$

heureka Dec 4, 2014
#2
+94558
0

Nice, heureka......the solution is easy.....figuring out how to get there is the hard part....!!!!!

Dec 4, 2014