Question

How would you solve these 2 maths problems?

Answer 1

Answer:

see explanation

Step-by-step explanation:

(9)

Expand and simplify right side and compare coefficients of like terms on left side.

(x - 2)(x² + ax + b) + c

= x³ + ax² + bx - 2x² - 2ax - 2b + c

= x³ + x²(a - 2) + x(b - 2a) - 2b + c

compare with

x³ + 2x² - 3x + 4

x² terms

a - 2 = 2 ( add 2 to both sides )

a = 4

x terms

b - 2a = - 3 ← substitute a = 4

b - 8 = - 3 ( add 8 to both sides )

b = 5

constant terms

- 2b + c = 4 ← substitute b = 5

- 10 + c = 4 ( add 10 to both sides )

c = 14

Thus a = 4, b = 5 and c = 14

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(10)

Since both functions cross the x- axis at - 2 then (- 2, 0) satisfies both, that is

f(- 2) = $(-2)^{4}$ + a(- 2)³ + b(- 2)² + 36(- 2) + 144 = 0, that is

- 16 - 8a + 4b - 72 + 144 = 0

- 8a + 4b + 56 = 0

- 8a + 4b = - 56 → (1)

and

g(- 2) = $(-2)^{4}$ + (a + 3)(- 2)³ - 23(- 2)² + (b + 10)(- 2) + 40 = 0, that is

- 16 - 8(a + 3) - 92 - 2(b + 10) + 40 = 0

- 16 - 8a - 24 - 92 - 2b - 20 + 40 = 0

- 8a - 2b - 112 = 0

- 8a - 2b = 112 → (2)

Subtract (1) from (2) term by term

- 6b = 168 ( divide both sides by - 6 )

b = - 28

Substitute b = - 28 into (2)

- 8a + 56 = 112 ( subtract 56 from both sides )

- 8a = 56 ( divide both sides by - 8 )

a = - 7

Thus a = - 7 and b = - 28