Answer:
y = -3 and y = -3
or y = -3 with a multiplicity of 2
Step-by-step explanation:
[tex](y+4)^2-2(y+4)+1[/tex]
If you want to factor this, heres what you do:
[tex](y+4)(y+4)-2(y+4)+1[/tex]
[tex]y^2+8y+16-2y-8+1[/tex]
[tex]y^2+6y+9[/tex]
Then factor:
[tex](y+3)(y+3)[/tex]
[tex](y+3)^2[/tex]
The value of y:
y + 3 = 0 and y + 3 = 0
y = -3 and y = -3
or y = -3 with a multiplicity of 2
The solution of the given function are; y = -3 and y = -3 or y = -3 with a multiplicity of 2.
A quadratic equation is the second-order degree algebraic expression in a variable. the standard form of this expression is ax² + bx + c = 0 where a. b are coefficients and x is the variable and c is a constant.
The given function is;
[tex](y + 4)^2 -2(y + 4) + 1[/tex]
To solve;
[tex](y + 4)(y + 4) -2(y + 4) + 1\\\\y^{2} + 8y + 16 - 2y -8 + 1\\\\y^{2} + 6y + 9 \\[/tex]
Then the factor are;
(y + 3)(y + 3)
The value of y:
y + 3 = 0 and y + 3 = 0
y = -3 and y = -3
or y = -3 with a multiplicity of 2
Learn more about quadratic equations;
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