1. Factor the expression $abx^3 + (a - 2b - ab)x^2 + (2b - a - 2)x + 2$ fully. Step 1: Group terms: $$abx^3 + (a - 2b - ab)x^2 + (2b - a - 2)x + 2 = (abx^3 - abx^2) + ((a - 2b)x^2) + ((2b - a - 2)x + 2)$$ Step 2: Factor within groups: - From $abx^3 - abx^2$, factor out $abx^2$: $$abx^2(x - 1)$$ - From $(a - 2b)x^2$, factor as is: $$(a - 2b)x^2$$ - For $(2b - a - 2)x + 2$, try to factor further. Step 3: Combine and attempt factorization by grouping: Rewrite: $$abx^3 + (a - 2b - ab)x^2 + (2b - a - 2)x + 2 = abx^3 - abx^2 + ax^2 - 2bx^2 + 2bx - ax - 2x + 2$$ Step 4: Group: $$ (abx^3 - abx^2) + (ax^2 - ax) + (-2bx^2 + 2bx) + (-2x + 2)$$ Step 5: Factor each group: - $abx^2(x - 1)$ - $a x (x - 1)$ - $-2b x (x - 1)$ - $-2 (x - 1)$ Step 6: Factor out common $(x - 1)$: $$ (x - 1)(abx^2 + a x - 2 b x - 2)$$ Step 7: Simplify the second factor: $$abx^2 + a x - 2 b x - 2 = abx^2 + x(a - 2b) - 2$$ Step 8: Check if it factors further: Try to factor as $(px + q)(rx + s)$: Expanding $(px + q)(rx + s) = pr x^2 + (ps + qr) x + qs$ Match coefficients: $$pr = ab$$ $$ps + qr = a - 2b$$ $$qs = -2$$ Try $q=2$, $s=-1$: $$qs=2 imes (-1) = -2$$ $$pr=ab$$ $$ps + qr = 2p - r = a - 2b$$ Try $p=b, r=a$: $$pr = ba = ab$$ correct. $$ps + qr = 2b - a$$ but should be $a - 2b$ which is negated. Try $q=-2$, $s=1$: $$qs = -2 imes 1 = -2$$ $$ps + qr = p(1) + q r = p - 2r = a - 2b$$ Try $p=a$, $r=b$: $$p r= a b = ab$$ correct. $$p - 2 r = a - 2 b$$ correct. Step 9: So factor is: $$ (a x - 2)(b x + 1)$$ Final factorization: $$ (x - 1)(a x - 2)(b x + 1)$$ 2. Find $a$ and $b$ such that the polynomial $x^4 + a x^3 + x^2 + b x + 1$ is divisible by $x^2 - 2 x + 1$. Step 1: Note that $x^2 - 2x + 1 = (x - 1)^2$. Step 2: Divisibility by $(x - 1)^2$ implies the polynomial and its first derivative vanish at $x=1$. Step 3: Evaluate the polynomial at $x=1$: $$1^4 + a (1)^3 + 1^2 + b (1) + 1 = 1 + a + 1 + b + 1 = a + b + 3 = 0$$ Step 4: Derivative is: $$4 x^3 + 3 a x^2 + 2 x + b$$ Evaluate at $x=1$: $$4 + 3 a + 2 + b = 6 + 3 a + b = 0$$ Step 5: Solve system: From step 3: $$a + b = -3$$ From step 4: $$3 a + b = -6$$ Subtract first from second: $$(3a + b) - (a + b) = -6 - (-3) \\ 2 a = -3 \\ a = -\frac{3}{2}$$ Step 6: Substitute $a$ back: $$-\frac{3}{2} + b = -3 \\ b = -3 + \frac{3}{2} = -\frac{3}{2}$$ Answer: $a = -\frac{3}{2}$, $b = -\frac{3}{2}$. 3. Find $a$ and $b$ given: $$ax^3 - 4x^2 + b x + 8$$ When divided by $(x - 2)$, remainder is $-8$. The curve passes through $(-1,7)$. Step 1: Use Remainder Theorem: Evaluate polynomial at $x=2$: $$a (2)^3 - 4 (2)^2 + b (2) + 8 = 8a - 16 + 2b + 8 = 8a + 2b - 8$$ Given remainder $= -8$, so: $$8a + 2b - 8 = -8$$ $$8a + 2b = 0$$ Step 2: Use point $(-1, 7)$: $$a (-1)^3 - 4 (-1)^2 + b (-1) + 8 = -a - 4 - b + 8 = -a - b + 4 = 7$$ Simplify: $$-a - b = 3$$ Step 3: Solve system: $$8a + 2b = 0$$ $$-a - b = 3$$ Multiply second by 2: $$-2a - 2b = 6$$ Add to first: $$(8a + 2b) + (-2a - 2b) = 0 + 6 \\ 6a = 6 \\ a = 1$$ Step 4: Find $b$: $$-1 - b = 3 \\ b = -4$$ Answer: $a=1$, $b=-4$.