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In the previous lesson, you learned how to factor polynomials of the form $x^2+bx+c$, and how to use that to solve equations that look like $x^2+bx+c=0$. In this lesson, you will learn how to do the same thing with quadratic polynomials where the $x^2$ term could have a coefficient different from 1.
The equation $y=a(2x^2-3x-2)$ is graphed on the grid to the left, with a slider for $a$.
As you saw in factorPict, multiplying the polynomial $2x^2-3x-2$ 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 $2x^2-6x+4=0$.
Check that these solutions solve the original equation $3x^2+12x-15$.
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 $-3x^2+2x+7$ 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 $x^2$ was $1$, we’ll start by looking more closely at the expansion of such products.
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 expand, when you expand $(px+q)(rx+s)$:
Say we want to write $3x^2-5x+2$ as a product of linear polynomials.
Let’s start by trying to factor $3x^2-5x+2$ 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 $3x^2-5x+2$. (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 slowFactor2, we know that if we can factor $8x^2+18x-5$, 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 $8x^2+18x-5$ 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.
Let’s look at the equation $-3x^2+4x-1=0$. Unlike the equations you’ve been solving, the coefficient of $x^2$ 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
As you saw in Questions factorPict and consFactor, multiplying a quadratic equation by a nonzero constant doesn’t change its solutions. What is the result of multiplying this equation by $3$?
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 $2x^2-13x+16$ from the left-hand side of the equation?
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.