Find the x and y Intercepts p(x)=(2x^2+7x+5)(x-3)
The question requests to find the x and y intercepts of the polynomial p(x) which is given as the product of two factors: a quadratic expression, \(2x^2+7x+5\), and a linear expression, \(x-3\).
The x-intercepts of a polynomial are the values of x for which the polynomial equals zero, meaning they are the solutions to the equation \(p(x) = 0\). Since p(x) is a product of two expressions, finding the x-intercepts involves determining the values of x where either \(2x^2+7x+5 = 0\) or \(x-3 = 0\).
The y-intercept of a polynomial is the value of the polynomial when x = 0, which is the constant term of the polynomial. To find the y-intercept, you simply evaluate p(x) at x = 0.
The question is asking for a solution to these two separate problems: finding the roots of the equation \(p(x) = 0\) to locate the x-intercepts, and evaluating the polynomial at x = 0 to determine the y-intercept.
$p \left(\right. x \left.\right) = \left(\right. 2 x^{2} + 7 x + 5 \left.\right) \left(\right. x - 3 \left.\right)$
Answer
Solution:
Step:1
Determine the x-intercepts.
Step:1.1
For x-intercepts, let $y = 0$ and solve the equation $0 = (2x^2 + 7x + 5)(x - 3)$.
Step:1.2
Proceed to solve the given equation.
Step:1.2.1
Express the equation as $(2x^2 + 7x + 5)(x - 3) = 0$.
Step:1.2.2
Recognize that if any factor equals $0$, the product is $0$: $2x^2 + 7x + 5 = 0$ and $x - 3 = 0$.
Step:1.2.3
Isolate $2x^2 + 7x + 5 = 0$ and find $x$.
Step:1.2.3.1
Set $2x^2 + 7x + 5$ to $0$.
Step:1.2.3.2
Resolve $2x^2 + 7x + 5 = 0$ for $x$.
Step:1.2.3.2.1
Employ grouping to factor.
Step:1.2.3.2.1.1
Decompose $7x$ into two terms that multiply to $2 \cdot 5 = 10$ and add to $7$.
Step:1.2.3.2.1.1.1
Extract $7$ from $7x$: $2x^2 + 7(x) + 5 = 0$.
Step:1.2.3.2.1.1.2
Rewrite $7$ as $2 + 5$: $2x^2 + (2 + 5)x + 5 = 0$.
Step:1.2.3.2.1.1.3
Apply distribution: $2x^2 + 2x + 5x + 5 = 0$.
Step:1.2.3.2.1.2
Factor out the common factors.
Step:1.2.3.2.1.2.1
Group terms: $(2x^2 + 2x) + (5x + 5) = 0$.
Step:1.2.3.2.1.2.2
Extract the GCF from each group: $2x(x + 1) + 5(x + 1) = 0$.
Step:1.2.3.2.1.3
Factor out the GCF, $x + 1$: $(x + 1)(2x + 5) = 0$.
Step:1.2.3.2.2
Each factor set to $0$ gives a solution: $x + 1 = 0$ and $2x + 5 = 0$.
Step:1.2.3.2.3
Solve $x + 1 = 0$.
Step:1.2.3.2.3.1
Set $x + 1$ to $0$.
Step:1.2.3.2.3.2
Subtract $1$ to find $x = -1$.
Step:1.2.3.2.4
Solve $2x + 5 = 0$.
Step:1.2.3.2.4.1
Set $2x + 5$ to $0$.
Step:1.2.3.2.4.2
Resolve $2x + 5 = 0$ for $x$.
Step:1.2.3.2.4.2.1
Subtract $5$: $2x = -5$.
Step:1.2.3.2.4.2.2
Divide by $2$: $x = \frac{-5}{2}$.
Step:1.2.3.2.5
The solutions are $x = -1, -\frac{5}{2}$.
Step:1.2.4
Solve $x - 3 = 0$.
Step:1.2.4.1
Set $x - 3$ to $0$.
Step:1.2.4.2
Add $3$ to find $x = 3$.
Step:1.2.5
The complete set of x-intercepts are $x = -1, -\frac{5}{2}, 3$.
Step:1.3
Express x-intercepts as points: $(-1, 0), \left(-\frac{5}{2}, 0\right), (3, 0)$.
Step:2
Identify the y-intercepts.
Step:2.1
For y-intercepts, set $x = 0$ and solve for $y$: $y = (2(0)^2 + 7(0) + 5)((0) - 3)$.
Step:2.2
Solve the resulting equation.
Step:2.2.1
Eliminate parentheses: $y = (2 \cdot 0^2 + 7 \cdot 0 + 5)(0 - 3)$.
Step:2.2.2
Further simplify: $y = (0 + 0 + 5)(-3)$.
Step:2.2.3
Finalize simplification: $y = 5 \cdot -3$.
Step:2.2.4
Calculate the result: $y = -15$.
Step:2.3
Express y-intercepts as points: $(0, -15)$.
Step:3
Compile the list of intercepts.
x-intercepts: $(-1, 0), \left(-\frac{5}{2}, 0\right), (3, 0)$ y-intercept: $(0, -15)$
Step:4
(No further action required in this step as it is empty in the original process.)
Knowledge Notes:
X-Intercepts: The points where a graph crosses the x-axis (y=0). To find them, set the function equal to zero and solve for x.
Y-Intercepts: The points where a graph crosses the y-axis (x=0). To find them, set x to zero and solve for y.
Factoring: A process used to simplify expressions and solve equations by expressing a polynomial as a product of its factors.
Factor by Grouping: A method used to factor polynomials with four terms by grouping pairs of terms that have a common factor.
Greatest Common Factor (GCF): The largest factor that divides two or more numbers. When factoring, the GCF is taken out to simplify the expression.
Quadratic Equations: Equations of the form $ax^2 + bx + c = 0$. They can be solved by factoring, completing the square, or using the quadratic formula.
Solving Linear Equations: To solve an equation like $ax + b = 0$, one would isolate x by subtracting b from both sides and then dividing by a.
