Factorize the following eqation x2+2x+15
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Answered by
2
Hey there !
Solution:
Equation: x² + 2x + 15
a = 1, b = 2, c = 15
=> Discriminant = b² - 4ac
=> ( 2 )² - 4 ( 1 ) ( 15 )
=> 4 - 60
=> - 56
Since - 56 < 0, the equation will contain imaginary or unreal roots.
Hence the equation cannot be factored with normal real roots.
Hope my answer helped !
Answered by
0
here is your answer OK
To solve this we want to factorise "x^2 -2x - 15" into the form (x+a)(x+b).
Now, if we expand (x+a)(x+b) we get x^2 + (a+b)x + ab. This means we have:
(x+a)(x+b) = x^2 + (a+b)x + ab = x^2 -2x - 15
Comparing these two equations we can see that we have (a+b)x = -2x and ab = -15. If we work through the factors of -15 (which are 1,-1,3,-3,5,-5,15,-15) we can see that we should pick a and b equal to 3 and -5.
So we have factorised the equation in the question to get:
x^2 -2x - 15 = (x+3)(x-5) = 0
Since any number times zero is equal to zero, we can see either x+3=0 or x-5=0. If we solve these equations we then get the solutions: x=-3 or x=5
To solve this we want to factorise "x^2 -2x - 15" into the form (x+a)(x+b).
Now, if we expand (x+a)(x+b) we get x^2 + (a+b)x + ab. This means we have:
(x+a)(x+b) = x^2 + (a+b)x + ab = x^2 -2x - 15
Comparing these two equations we can see that we have (a+b)x = -2x and ab = -15. If we work through the factors of -15 (which are 1,-1,3,-3,5,-5,15,-15) we can see that we should pick a and b equal to 3 and -5.
So we have factorised the equation in the question to get:
x^2 -2x - 15 = (x+3)(x-5) = 0
Since any number times zero is equal to zero, we can see either x+3=0 or x-5=0. If we solve these equations we then get the solutions: x=-3 or x=5
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