Factories the following quadratic equations 10Y2 + 21Y-10
10y²+21y-10=0
10y²+25y-4y-10=0
(10y²+25)-(4y+10)=0
5y(2y+5)-2(2y+5)=0
(5y-2)(2y+5)=0
5y-2=0 or 2y+5=0
Y=2/5 or y=-5/2
typo?
(10y²+25)-(4y+10)=0
or
(10y²+25y)-(4y+10)=0
Thanks sir yes it was a typo
To factorize the quadratic equation 10Y^2 + 21Y - 10, we need to find two binomials that multiply together to give this quadratic expression. The general form of a quadratic equation is ax^2 + bx + c, where a, b, and c are constants.
In this case, a = 10, b = 21, and c = -10. To factorize, we need to find two numbers that multiply together to give ac (product of the leading coefficient and the constant term) and add up to b (the coefficient of the middle term).
The product of ac is (10)(-10) = -100, and we need to find two numbers whose sum is 21. Upon inspection, we can see that the numbers are 25 and -4 because 25 * -4 = -100 and 25 + (-4) = 21.
Next, we rewrite the middle term (21Y) using these two numbers:
10Y^2 + 25Y - 4Y - 10
Now, we group the terms into two pairs:
(10Y^2 + 25Y) + (-4Y - 10)
We can factor out the greatest common factor (GCF) from each pair:
5Y(2Y + 5) - 2(2Y + 5)
Notice that we have a common binomial factor, (2Y + 5), which can be factored out:
(2Y + 5)(5Y - 2)
Therefore, the factored form of the quadratic equation 10Y^2 + 21Y - 10 is (2Y + 5)(5Y - 2).