Page 159 - Algebra

Explanation:
1. When (x – 3) is a factor of f(x) = b(x4 – 3x2) + c(x3 f(3) = 0
b(34 –3.32)+c(33 –9)=0 54b + 18c = 0
3b + c = 0
2. Group the terms
x4 –7x3 +4x–28=0 x3(x – 7) + 4(x – 7) = 0 (x3+ 4)(x – 7) = 0
x3 = – 4 or x = 7
x3 = – 4 is not a real value
x = 7 is the solution of the equation
– 9)
3. When polynomial 3x3 – 8x2 + 5x – 18 is divided by x – 3, the remainder is p(3) =3.33 –8.32 +5.3–18
= 81 – 72 + 15 – 18
=6
4.
5. f(x) =x4 – mx2 – 10x + 24
The factors of the polynomial intersecting the x-axis at (4, 0) (-3, 0) (1, 0) and (b, 0)is: f(x) = (x – 4)(x + 3) (x – 1) (x – b)
When you multiply the factors of the function, the result is the constant term. (–4)(3)( –1)( –b) = 24
– 12b = 24
b = –2
f(x) = (x – 4)(x + 3) (x – 1) (x + 2)
=(x2 +3x–4x–12)(x2 +2x–x–2)
=(x2 –x–12)(x2 +x–2)
= x4 + x3 – 2x2 – x3 – x2 + 2x – 12x2 – 12x + 24 = x4 – 15x2 – 10x + 24
Compare with the original function.
m = 15
6. Here, you have to find the roots of the equation. f(x) = 0
9x4 – 12x3 – 6x + 8 = 0 3x3(3x – 4) – 2(3x – 4) = 0 (3x3– 2)(3x – 4)
x =4 3
All the other three roots are not real and will not intersect the x-axis. Therefore, the factors of equation intersect only at one point.
A polynomial function f(x) has (x –a) as a factor only if f(a) = 0. So, x + 6 is a factor of the
polynomial function.
Page 158 of 177
Algebra I & II