Question

lesson 1.4 practice b for use with the lesson \solve ax²+bx + c = 0 by factoring\ factor the expression. if the expression cannot be factored, say so. 1. 3x²+10x - 8 2. 2x²+5x - 3 3. 4x²+4x + 1 4. 2x²-5x + 1 5. 4x²+5x - 6 6. 2x²+11x + 15 7. 9x²+12x + 4 8. 12x²-24x + 9 9. 18x²-2 10. 12x²+17x + 6 11. 15x²+8x - 16 12. 4x²-5 13. 12x²-39x + 9 14. 18x²-9x - 14 15. 20x²-54x + 36 16. 42x²+35x + 7 17. -12x²-x + 11 18. 80x²+68x + 12 solve the equation. 19. 2x²+3x - 2 = 0 20. 2x²-3x - 9 = 0 21. 4x²-8x + 3 = 0 22. 9x²-4 = 0 23. 8x²-6x + 1 = 0 24. 18x²+48x=-32 25. 9x²+11x + 18=-10x + 8 26. 5x²-2x - 6=-3x²+6x 27. 5x²-3x + 3=-2x²+3 28. 25x²-24x - 9=-7x²+12x - 18 find the zeros of the function by rewriting the function in intercept form. 29. y = 3x²+2x 30. y = 12x²+8x - 15 31. f(x)=5x²-25x + 30 32. y = 25x²+10x - 24 33. g(x)=33x²-9x - 24 34. y = 4x²+1 find the value of x. 35. area of the triangle = 27 36. area of the rectangle = 22 37. picture frame you are making a frame of uniform width for a picture that is to be displayed at the local museum. the picture is 3.25 feet tall and 3 feet wide. the museum has allocated 15 square feet of wall space to display the picture. what should the width of the frame be in order to use all of the allocated space?

Explanation:

Step1: Factor $3x^{2}+10x - 8$

We need to find two numbers that multiply to $3\times(-8)=- 24$ and add up to $10$. The numbers are $12$ and $-2$.
$3x^{2}+10x - 8=3x^{2}+12x-2x - 8=3x(x + 4)-2(x + 4)=(3x - 2)(x + 4)$

Step2: Solve $2x^{2}+3x - 2 = 0$

Factor the left - hand side. We find two numbers that multiply to $2\times(-2)=-4$ and add up to $3$. The numbers are $4$ and $-1$.
$2x^{2}+3x - 2=2x^{2}+4x-x - 2=2x(x + 2)-(x + 2)=(2x - 1)(x + 2)=0$
Then, using the zero - product property $2x-1 = 0$ or $x + 2=0$.
If $2x-1=0$, then $2x=1$, $x=\frac{1}{2}$; if $x + 2=0$, then $x=-2$.

Step3: Find the zeros of $y = 3x^{2}+2x$

Factor out the greatest common factor $x$.
$y=x(3x + 2)$
Set $y = 0$, then $x=0$ or $3x+2 = 0$. If $3x+2 = 0$, then $3x=-2$, $x=-\frac{2}{3}$.

Step4: Solve for $x$ in the triangle area problem

The area of a triangle is $A=\frac{1}{2}\times base\times height$. Here, $A = 27$, base $=4x + 1$, and height $=3x$.
So, $\frac{1}{2}(4x + 1)\times3x=27$.
Multiply both sides by $2$ to get $(4x + 1)\times3x=54$.
Expand to $12x^{2}+3x-54 = 0$. Divide through by $3$: $4x^{2}+x - 18=0$.
Factor: $4x^{2}+x - 18=(4x + 9)(x - 2)=0$.
So, $x = 2$ or $x=-\frac{9}{4}$. Since $x$ represents a length, we discard $x=-\frac{9}{4}$.

Step5: Solve for $x$ in the rectangle area problem

The area of a rectangle is $A = length\times width$. Here, $A = 22$, length $=3x + 1$, and width $=2x+1$.
So, $(3x + 1)(2x + 1)=22$.
Expand: $6x^{2}+3x+2x + 1=22$, or $6x^{2}+5x-21 = 0$.
Factor: $6x^{2}+5x-21=(2x - 3)(3x+7)=0$.
So, $x=\frac{3}{2}$ or $x=-\frac{7}{3}$. Since $x$ represents a length, we discard $x=-\frac{7}{3}$.

Step6: Solve for $x$ in the picture - frame problem

The area of the picture and frame together is $(3 + 2x)(3.25+2x)$.
We know that $(3 + 2x)(3.25+2x)=15$.
Expand: $9.75+6x+6.5x + 4x^{2}=15$.
Rearrange to $4x^{2}+12.5x-5.25 = 0$. Multiply through by $4$ to get $16x^{2}+50x-21 = 0$.
Factor: $16x^{2}+50x-21=(8x - 3)(2x+7)=0$.
So, $x=\frac{3}{8}=0.375$ or $x=-\frac{7}{2}$. Since $x$ represents a width, we discard $x=-\frac{7}{2}$.

Answer:

1. $(3x - 2)(x + 4)$
2. $x=\frac{1}{2},x=-2$
3. $x = 0,x=-\frac{2}{3}$
4. $x = 2$
5. $x=\frac{3}{2}$
6. $x = 0.375$ feet