Solve 70y+49+25y^2=

$$70 y + 49 + 25 y ^ { 2 } =$$

Factor

$$\left(5y+7\right)^{2}$$

Solution Steps

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$25y^{2}+70y+49$$

Factor the expression by grouping. First, the expression needs to be rewritten as 25y^{2}+ay+by+49. To find a and b, set up a system to be solved.

$$a+b=70 ab=25\times 49=1225$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 1225.

$$1,1225 5,245 7,175 25,49 35,35$$

Calculate the sum for each pair.

$$1+1225=1226 5+245=250 7+175=182 25+49=74 35+35=70$$

The solution is the pair that gives sum 70.

$$a=35 b=35$$

Rewrite 25y^{2}+70y+49 as \left(25y^{2}+35y\right)+\left(35y+49\right).

$$\left(25y^{2}+35y\right)+\left(35y+49\right)$$

Factor out 5y in the first and 7 in the second group.

$$5y\left(5y+7\right)+7\left(5y+7\right)$$

Factor out common term 5y+7 by using distributive property.

$$\left(5y+7\right)\left(5y+7\right)$$

Rewrite as a binomial square.

$$\left(5y+7\right)^{2}$$

Steps Using Square Of Binomial

This trinomial has the form of a trinomial square, perhaps multiplied by a common factor. Trinomial squares can be factored by finding the square roots of the leading and trailing terms.

$$factor(25y^{2}+70y+49)$$

Find the greatest common factor of the coefficients.

$$gcf(25,70,49)=1$$

Find the square root of the leading term, 25y^{2}.

$$\sqrt{25y^{2}}=5y$$

Find the square root of the trailing term, 49.

$$\sqrt{49}=7$$

The trinomial square is the square of the binomial that is the sum or difference of the square roots of the leading and trailing terms, with the sign determined by the sign of the middle term of the trinomial square.

$$\left(5y+7\right)^{2}$$

Steps Using the Quadratic Formula

Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

$$25y^{2}+70y+49=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$y=\frac{-70±\sqrt{70^{2}-4\times 25\times 49}}{2\times 25}$$

Square 70.

$$y=\frac{-70±\sqrt{4900-4\times 25\times 49}}{2\times 25}$$

Multiply -4 times 25.

$$y=\frac{-70±\sqrt{4900-100\times 49}}{2\times 25}$$

Multiply -100 times 49.

$$y=\frac{-70±\sqrt{4900-4900}}{2\times 25}$$

Add 4900 to -4900.

$$y=\frac{-70±\sqrt{0}}{2\times 25}$$

Take the square root of 0.

$$y=\frac{-70±0}{2\times 25}$$

Multiply 2 times 25.

$$y=\frac{-70±0}{50}$$

Factor the original expression using ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right). Substitute -\frac{7}{5} for x_{1} and -\frac{7}{5} for x_{2}.

$$25y^{2}+70y+49=25\left(y-\left(-\frac{7}{5}\right)\right)\left(y-\left(-\frac{7}{5}\right)\right)$$

Simplify all the expressions of the form p-\left(-q\right) to p+q.

$$25y^{2}+70y+49=25\left(y+\frac{7}{5}\right)\left(y+\frac{7}{5}\right)$$

Add \frac{7}{5} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25y^{2}+70y+49=25\times \frac{5y+7}{5}\left(y+\frac{7}{5}\right)$$

Add \frac{7}{5} to y by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$25y^{2}+70y+49=25\times \frac{5y+7}{5}\times \frac{5y+7}{5}$$

Multiply \frac{5y+7}{5} times \frac{5y+7}{5} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$25y^{2}+70y+49=25\times \frac{\left(5y+7\right)\left(5y+7\right)}{5\times 5}$$

Multiply 5 times 5.

$$25y^{2}+70y+49=25\times \frac{\left(5y+7\right)\left(5y+7\right)}{25}$$

Cancel out 25, the greatest common factor in 25 and 25.

$$25y^{2}+70y+49=\left(5y+7\right)\left(5y+7\right)$$

Evaluate

$$\left(5y+7\right)^{2}$$