factorise 6x^3+10x^2+8x
Answers
Answer:
f (x)=6x^3+10x^2-8x-8
f (1)=0 divide f (x) by (x-1) to get
6x^2+16x+8=2 (3x^2+8x+4)=2 (3x+2)(x+2)
So f (x)=2 (x-1)(3x+2)(x+2)
First, divide by 2. You now have
3x^3 + 5x^2 - 4x - 4.
If it has a factor ax + b, where a and b are integers, then a must be 1, 3, -1, or -3. Also, b must be 1, 2, 4, -1, -2, -4. In fact we can assume a is positive, so it must 1 or 3. That gives us 12 possibilities (2 for a, and 6 for b). Try dividing each of those 12 linear factors into 3x^3 + 5x^2 - 4x -4.
If the remainder of this division is 0, you now have a product of a linear and a quadratic.
Repeat this process with the quadratic, etc.
Let f(x) = 2(3x³+5x²-4x-4)
f(1)=0→(x-1) is a factor
∴ f(x)=2(3x²+8x +4)(x-1)
=2(3x+2)(x+2)(x+1)
Factor 6x3+10x2−8x−8
6x3+10x2−8x−8
=2(x−1)(x+2)(3x+2)
Answer:
2(x−1)(x+2)(3x+2)
2X(3x²+5x-8)
2X(3x²+8x-3x-8)
2X(x-1)(3x+8)
X=1,-16/3
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