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In the previous lesson, you learned how to factor polynomials of the form x2+bx+c, and how to use that to solve equations that look like x2+bx+c=0. In this lesson, you will learn how to do the same thing with quadratic polynomials where the x2 term could have a coefficient different from 1.
Question 1 of 13. The equation y=a(2x2−3x−2) is graphed on the grid to the left, with a slider for a.
As you saw in Question 1, multiplying the polynomial 2x2−3x−2 by a nonzero constant doesn’t change when that polynomial is zero. In fact, you can do this for any polynomial, because ab=0 exactly when a=0 or b=0. This can be useful in solving equations.
Check that these solutions solve the original equation 2x2−6x+4=0.
Check that these solutions solve the original equation 3x2+12x−15.
Question 3. Write each quadratic polynomial in the table below as the product of a number and a simpler quadratic polynomial. Use the largest possible number.
Write the polynomial −3x2+2x+7 as the product of of −1 and another quadratic polynomial.
In order to solve quadratic equations, we want to be able to write quadratic polynomials as the product of linear polynomials. Just as we did when the coefficient of x2 was 1, we’ll start by looking more closely at the expansion of such products.
Question 4. Expand each product (px+q)(rx+s) below, using scratch paper. Note the values of pr and qs.
Look at the second and third rows of the table. Notice that in the second row we’re multiplying (3x+2) by (2x+5), while in the third row we’re multiplying (−3x−2), which is the opposite of (3x+2), by (−2x−5), which is the opposite of (2x+5).
As you saw in Question 4, when you expand (px+q)(rx+s):
Question 5. Say we want to write 3x2−5x+2 as a product of linear polynomials.
Let’s start by trying to factor 3x2−5x+2 as (3x+q)(x+s) with qs=2.
Write (3x+q)(x+s) with qs=2 in all possible ways, and then expand those products to see if they equal 3x2−5x+2. (The pairs of integers you just found could be q and s in either order.)
If you have a way of factoring a quadratic polynomial into two linear polynomials with negative linear coefficients, the factors will be the opposites of a way of factoring that polynomial into linear polynomials with positive linear coefficients. As a result:
In order to find out if a quadratic polynomial with a positive quadratic coefficient can be factored into linear polynomials, you only need to check whether it can be factored into linear polynomials with positive linear coefficients.
In the next few questions, we’ll solve the equation
by factoring its left-hand side.
From Question 7, we know that if we can factor 8x2+18x−5, we can factor it as either (8x+q)(x+s) or (4x+q)(2x+s), where qs=−5.
We’ll start by trying to factor the polynomial as (8x+q)(x+s). Write (8x+q)(x+s) with qs=−5 in all possible ways, and then expand those products.
Now, we’ll try factoring it as (4x+q)(2x+s). Write (4x+q)(2x+s) with qs=−5 in all possible ways, and then expand those products.
Using your work from the two tables above, write 8x2+18x−5 as the product of two linear polynomials.
In the last few questions, you learned how to solve certain quadratic equations. In this section, we’ll look at some ways of turning some other quadratic equations into ones you already know how to solve.
Question 9. Let’s look at the equation −3x2+4x−1=0. Unlike the equations you’ve been solving, the coefficient of x2 here is negative.
What is the result of multiplying this equation by −1?
What about quadratic equations that involve fractions? For example, let’s look at how you would solve the equation
Question 10. As you saw in Questions 1 and 2, multiplying a quadratic equation by a nonzero constant doesn’t change its solutions. What is the result of multiplying this equation by 3?
Question 11. Each equation in the table below includes a polynomial with fractional coefficients. Multiply both sides of that equation by a constant, to get an equation where all the coefficients are integers. Use the smallest possible positive constant. You do not have to solve the resulting equation.
Now, say you wanted to solve an equation like
in which two different polynomials are set equal to each other.
What is the result of subtracting 2x2−13x+16 from the left-hand side of the equation?
Question 13. Each row of the table below has an equation involving two quadratic polynomials. By subtracting the same quantity from both sides, rewrite those equations as equations involving only one quadratic polynomial set equal to 0. Use scratch paper.