If (x^ 2 + 5x + 5)x + 5 = 1 and then the number of integers...

Given that,& \left(x^{2} + 5x + 5\right)x + 5 = 1$</p><p>This is possible when~$x^{2} + 5x + 5 \ne 0 \text{ and } x + 5 = 0$~or~$x^{2} + 5x + 5 = 1$~or~$x^{2} + 5x + 5 = - 1 \text{ and } x + 5 = eveninteger$</p><p>Case1,~$x + 5 = 0$</p><p>$⇒ x = - 5$</p><p>And for this value of~$x$,~$x^{2} + 5x + 5 \ne 0$</p><p>Case2,~$x^{2} + 5x + 5 = 1$</p><p>$⇒ x^{2} + 5x + 4 = 0$</p><p>$⇒ \left(x + 1\right)\left(x + 4\right) = 0$</p><p>$⇒ x = - 1, - 4$</p><p>Case3,~$x^{2} + 5x + 5 = - 1$</p><p>$⇒ x^{2} + 5x + 6 = 0$</p><p>$⇒ \left(x + 2\right)\left(x + 3\right) = 0$</p><p>$⇒ x = - 2, - 3$</p><p>But for~$x = - 2,$~we get~$x + 5 = 3$~which is not even.</p><p>Hence, the number of integers satisfying this equation is~$4$.</p>$