The question is ax^2+bx+c=(2x-7)(x-3) for all values of x, What is the value of b?
The teacher went over and said the answer is -13, but I am not sure how she got that.
Can you please explain.
Thank you!
I did the first part
(2x-7)(x-3)
2x^2-6x-7x+21
2x^2-13x+21
here x=-13 but how do you find out the value of b?
Your teacher wanted you to match up the corresponding terms.
compare
ax^2 + bx + c with
2x^2 - 13x + 21 , and it says they are equal
so 2 matches with a, -13 matches with b, and
21 matches with c
If two polynomials are identical, all the coefficients must match. So,
a = 2
b = -13
c = 21
Thank you.
I was getting thrown off since the bx is +bx and since x=-13 I wasn't sure if you would change the sign to +13 since the bx was positive.
To find the value of b, we need to compare the given quadratic equation ax^2 + bx + c with the factored form (2x - 7)(x - 3).
In the factored form, we have two binomials multiplied together: (2x - 7) and (x - 3). To expand this, we multiply each term in the first binomial by each term in the second binomial:
(2x)(x) + (2x)(-3) + (-7)(x) + (-7)(-3)
This simplifies to:
2x^2 - 6x - 7x + 21
Combining like terms:
2x^2 - 13x + 21
Comparing this to the given equation ax^2 + bx + c, we can see that a = 2, b = -13, and c = 21. Therefore, the value of b is indeed -13, as your teacher stated.
So, the value of b in the given equation ax^2 + bx + c = (2x - 7)(x - 3) for all values of x is -13.