To find the highest common factor (HCF) of the given expressions using the division method, we divide the first expression by the second expression and continue the process until the remainder becomes zero.
y³ - 3y + 2 ÷ y³ - 5y² + 7y - 3
We need to divide the highest degree term of the dividend by the highest degree term of the divisor, which gives us:
y³ ÷ y³ = 1
Now, we multiply the entire divisor by this quotient, which gives us:
y³ - 5y² + 7y - 3
Next, we subtract this from the dividend, which gives us:
-2y² + 7y - 1
Now, we repeat the process by dividing the highest degree term of the divisor (y³) by the highest degree term of the remainder (-2y²), which gives us:
y³ ÷ -2y² = -y
We then multiply the entire divisor by this quotient, which gives us:
-y³ + 5y² - 7y + 3
And subtracting this from the remainder, we get:
3y² - 6y + 4
We repeat this process until we get a remainder of zero:
3y² - 6y + 4 ÷ -y
= -3y² + 6y - 4
-3y² + 6y - 4 ÷ -3
= y² - 2y + 1
y² - 2y + 1 ÷ y - 1
= y - 1To determine the highest common factor (HCF) of the provided expressions using the division technique, split the first expression by the second expression and repeat the process until the residual becomes zero.
y³ - 3y + 2 ÷ y³ - 5y² + 7y - 3
We must split the highest degree term of the payout by the highest degree term of the divisor, yielding:
y³ ÷ y³ = 1
Now we multiply the full fraction by this quotient, which gives us:
y³ - 5y² + 7y - 3
Subtracting this from the payout yields:
-2y² + 7y - 1
Now we replicate the procedure by dividing the highest degree term of the divisor (y3) by the highest degree term of the remainder (-2y2), yielding:
y³ ÷ -2y² = -y
The full fraction is then multiplied by this quotient, yielding:
-y³ + 5y² - 7y + 3
And when we deduct this from the rest, we get:
3y² - 6y + 4
This procedure is repeated until we get a residual of zero:
3y² - 6y + 4 ÷ -y
= -3y² + 6y - 4
-3y² + 6y - 4 ÷ -3
= y² - 2y + 1
y² - 2y + 1 ÷ y - 1
= y - 1
As a result, the HCF for y3 - 3y + 2 and y3 - 5y2 + 7y - 3 is y - 1.