QuestionFactor the polynomial $10 x^{2}+29 x+10$ as $(5 x+B)(C x+D)$ with integers $B$ , $C$ , and $D$ . Find $B$ , $C$ , and $D$ .

Studdy Solution

STEP 1

Assumptions1. The polynomial is $10x^ +29x +10$ . The polynomial can be factored into the form $(5x + B)(Cx + D)$ where B, C, and D are integers

STEP 2

The general form of a quadratic equation is $ax^2 + bx + c$ . In this case, $a =10$ , $b =29$ , and $c =10$ . We need to find two numbers that multiply to $ac =10 \times10 =100$ and add to $b =29$ .

STEP 3

The two numbers that satisfy these conditions are20 and5, because $20 \times5 =100$ and $20 +5 =25$ .

STEP 4

Now we rewrite the middle term of the quadratic equation as the sum of the terms $20x$ and $x$ . $10x^2 +29x +10 =10x^2 +20x +9x +10$

STEP 5

Next, we factor by grouping. This means we group the first two terms together and the last two terms together, and factor out the greatest common factor from each group.
$10x^2 +20x +9x +10 =10x(x +2) +5(2x +2)$

STEP 6

Now we can see that $(x +2)$ is a common factor in both terms. We factor this out to get the final factored form of the quadratic equation.
$10x(x +2) +5(2x +2) = (5x +5)(2x +2)$ So, the polynomial $10x^2 +29x +10$ factors to $(5x +5)(2x +2)$ .
Therefore, $B =5$ , $C =2$ , and $=2$ .

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QuestionFactor the polynomial $10 x^{2}+29 x+10$ as $(5 x+B)(C x+D)$ with integers $B$ , $C$ , and $D$ . Find $B$ , $C$ , and $D$ .

Studdy Solution

STEP 1

Assumptions1. The polynomial is $10x^ +29x +10$ . The polynomial can be factored into the form $(5x + B)(Cx + D)$ where B, C, and D are integers

STEP 2

The general form of a quadratic equation is $ax^2 + bx + c$ . In this case, $a =10$ , $b =29$ , and $c =10$ . We need to find two numbers that multiply to $ac =10 \times10 =100$ and add to $b =29$ .

STEP 3

The two numbers that satisfy these conditions are20 and5, because $20 \times5 =100$ and $20 +5 =25$ .

STEP 4

Now we rewrite the middle term of the quadratic equation as the sum of the terms $20x$ and $x$ . $10x^2 +29x +10 =10x^2 +20x +9x +10$

STEP 5

Next, we factor by grouping. This means we group the first two terms together and the last two terms together, and factor out the greatest common factor from each group.
$10x^2 +20x +9x +10 =10x(x +2) +5(2x +2)$

STEP 6

Now we can see that $(x +2)$ is a common factor in both terms. We factor this out to get the final factored form of the quadratic equation.
$10x(x +2) +5(2x +2) = (5x +5)(2x +2)$ So, the polynomial $10x^2 +29x +10$ factors to $(5x +5)(2x +2)$ .
Therefore, $B =5$ , $C =2$ , and $=2$ .

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QuestionFactor the polynomial 10x2+29x+1010 x^{2}+29 x+1010x2+29x+10 as (5x+B)(Cx+D)(5 x+B)(C x+D)(5x+B)(Cx+D) with integers BBB, CCC, and DDD. Find BBB, CCC, and DDD.

Studdy Solution

STEP 1

Assumptions1. The polynomial is 10x+29x+1010x^ +29x +1010x+29x+10 . The polynomial can be factored into the form (5x+B)(Cx+D)(5x + B)(Cx + D)(5x+B)(Cx+D) where B, C, and D are integers

STEP 2

The general form of a quadratic equation is ax2+bx+cax^2 + bx + cax2+bx+c. In this case, a=10a =10a=10, b=29b =29b=29, and c=10c =10c=10. We need to find two numbers that multiply to ac=10×10=100ac =10 \times10 =100ac=10×10=100 and add to b=29b =29b=29.

STEP 3

The two numbers that satisfy these conditions are20 and5, because 20×5=10020 \times5 =10020×5=100 and 20+5=2520 +5 =2520+5=25.

STEP 4

Now we rewrite the middle term of the quadratic equation as the sum of the terms 20x20x20x and xxx.10x2+29x+10=10x2+20x+9x+1010x^2 +29x +10 =10x^2 +20x +9x +1010x2+29x+10=10x2+20x+9x+10

STEP 5

Next, we factor by grouping. This means we group the first two terms together and the last two terms together, and factor out the greatest common factor from each group.10x2+20x+9x+10=10x(x+2)+5(2x+2)10x^2 +20x +9x +10 =10x(x +2) +5(2x +2)10x2+20x+9x+10=10x(x+2)+5(2x+2)

STEP 6

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QuestionFactor the polynomial $10 x^{2}+29 x+10$ as $(5 x+B)(C x+D)$ with integers $B$ , $C$ , and $D$ . Find $B$ , $C$ , and $D$ .

Assumptions1. The polynomial is $10x^ +29x +10$ . The polynomial can be factored into the form $(5x + B)(Cx + D)$ where B, C, and D are integers

The general form of a quadratic equation is $ax^2 + bx + c$ . In this case, $a =10$ , $b =29$ , and $c =10$ . We need to find two numbers that multiply to $ac =10 \times10 =100$ and add to $b =29$ .

The two numbers that satisfy these conditions are20 and5, because $20 \times5 =100$ and $20 +5 =25$ .

Now we rewrite the middle term of the quadratic equation as the sum of the terms $20x$ and $x$ . $10x^2 +29x +10 =10x^2 +20x +9x +10$

Next, we factor by grouping. This means we group the first two terms together and the last two terms together, and factor out the greatest common factor from each group.
$10x^2 +20x +9x +10 =10x(x +2) +5(2x +2)$