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You can do this by finding two numbers whose product is equal to the product of the coefficients on the first and last terms, and whose sum is equal to the coefficient of the middle term. So we want two numbers, call them A and B, that add to make 3 and multiply to make -2 * 9, or -18. Those numbers would be six and negative three.
You can take those numbers and break the middle term down into two terms using them as their coefficients:
9x2 + 3x - 2
= 9x2 + 6x - 3x - 2
Now factor out like terms:
= 3x(3x + 2) - 1(3x + 2)
And group:
= (3x - 1)(3x + 2)
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How do you factor 18X squared plus 2?
18x2 + 2= 2(9x2 + 1)= 2(9x2 - -1)= 2[(3x)2 - i2]= 2(3x + √-1)(3x - √-1) (substitute i for √-1)= 2(3x + i)(3x - i)
9x2 plus 12x plus 4 factored in decending order?
9x2...I guess that means 9x squared ? The problem: 9x(squared)+12x+4 Here's the answer: (3x+2)(3x+2) or even more simplified: (3x+2) squared
What is 3xto the power of 2 -4?
(3x)^(2-4) = (3x)^-2 = (9x2)^-1 = 1/(9x2)
How do you factorize 9 multiplied by x square plus 24 multiplied by x and y plus 16 multiplied by y square?
9x2 + 24xy + 16y2 = 9x2 + 12xy + 12xy + 16y2 = 3x(3x + 4y) + 4y(3x + 4y) = (3x + 4y)*(3x + 4y) = (3x + 4y)2
How can the factoring of the first equation help you factor the second equation 9x2 power 18x plus 8 9a2power 18ab plus 8b2power?
The first equation factors to (3x + 2)(3x + 4) The second equation factors to (3a + 2b)(3a + 4b)
Factor 9x2 plus 12x plus 4?
9x2 + 12x + 4 = 9x2 + 6x + 6x + 4 = 3x(3x + 2) + 2(3x + 2) = (3x + 2)(3x + 2) = (3x + 2)2
How do you factor the given polynomial 9x2 plus 9x-10?
9x2 + 9x - 10 =(3x - 2)(3x + 5)
How do you factor 9x squared plus 6x plus 1?
9x2 + 6x + 1 = 9x2 + 3x + 3x + 1 = 3x(3x + 1) + 1(3x + 1) = (3x + 1)(3x + 1) or (3x +1)2
How do you factor 18X squared plus 2?
18x2 + 2= 2(9x2 + 1)= 2(9x2 - -1)= 2[(3x)2 - i2]= 2(3x + √-1)(3x - √-1) (substitute i for √-1)= 2(3x + i)(3x - i)
How do you factor 3x plus 8b times 9x2 - 24xb plus 64b2?
One of the factors is 3x + 8b. The other is 9x^2 -24xb + 64b^2
What is the main factor of 3x3-9x2 3x?
3x(x^2 - 3x + 1)
How would you factor 9x2 - 6xy plus y2 - 81?
9x^2-6xy+y^2 = (y-3x)^2 (y-3x)^2-81 -> ((y-3x)+9)((y-3x)-9)
How do you factor -6x plus 9x2-24?
-6x + 9x2 - 24 = 3(3x2 - 2x - 8) = 3(3x2 - 6x + 4x - 8) = 3(3x(x - 2) + 4(x - 2)) = 3(3x + 4)(x - 2)
What are the factors of 9x2 3x-2?
If you mean: 9x2+3x-2 = (3x+2)(3x-1) when factored
What is the factor of 9x2 minus 12x minus 4?
There's a chance this was notated incorrectly. It looks like it wants to be some combination of 3x and 2. 9x2 + 12x + 4 = (3x + 2)(3x + 2) or (3x + 2)2 -9x2 - 12x - 4 = -(3x + 2)2 9x2 - 12x + 4 = (3x - 2)2 -9x2 + 12x - 4 = -(3x - 2)2
9x2 plus 12x plus 4 factored in decending order?
9x2...I guess that means 9x squared ? The problem: 9x(squared)+12x+4 Here's the answer: (3x+2)(3x+2) or even more simplified: (3x+2) squared
How do you factor 9x2-6xy plus y2-81?
9x2 - 6xy + y2 - 81 = (3x - y)2 - 81 (since the first three terms make a perfect square = (3x - y - 9)(3x - y + 9) since the expression now is a difference of two squares.
