Q.(a) Solve the equation $(x+1)(x+3)(x-2)(x-4)+21=0$ OR

(a) Pairs the four linear factors so both products share the quadratic $x^2-x$, substitutes to reduce to a quadratic in that expression, and solves; (b) analyses domain, intercepts, asymptote, monotonicity and concavity of $y=\log(1+x)$ to describe its sketch. Both alternatives answered below.

(a) Solve $(x+1)(x+3)(x-2)(x-4)+21=0$

1. Pair the factors to share a common quadratic.

$(x+1)(x-2)=x^2-x-2,\qquad(x+3)(x-4)=x^2-x-12$

Both contain $x^2-x$.

2. Substitute $t=x^2-x$. The equation becomes

$(t-2)(t-12)+21=0$

3. Expand.

$t^2-14t+24+21=0\ \Longrightarrow\ t^2-14t+45=0$

4. Solve the quadratic in $t$.

$t=\dfrac{14\pm\sqrt{196-180}}2=\dfrac{14\pm\sqrt{16}}2=\dfrac{14\pm4}2=9\ \text{or}\ 5$

5. Case $t=9$: solve $x^2-x-9=0$.

$x=\dfrac{1\pm\sqrt{1+36}}2=\dfrac{1\pm\sqrt{37}}2$

6. Case $t=5$: solve $x^2-x-5=0$.

$x=\dfrac{1\pm\sqrt{1+20}}2=\dfrac{1\pm\sqrt{21}}2$

7. All four roots: $x=\dfrac{1\pm\sqrt{37}}2,\ \dfrac{1\pm\sqrt{21}}2$.

(b) Sketch the curve $y=\log(1+x)$

1. Domain. Need $1+x>0\Rightarrow x>-1$.

2. Vertical asymptote. As $x\to-1^+$, $1+x\to0^+$, so $y=\log(1+x)\to-\infty$. The line $x=-1$ is a vertical asymptote.

3. Intercepts. At $x=0$: $y=\log1=0$, so the curve passes through the origin $(0,0)$ — this is the only intercept (the curve never touches $x=-1$).

4. Monotonicity. $y'=\dfrac1{1+x}>0$ for all $x>-1$, so the curve is strictly increasing throughout its domain.

5. Concavity. $y''=-\dfrac1{(1+x)^2}<0$ for all $x>-1$, so the curve is concave down (convex upward) everywhere — no points of inflection.

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